


.. . ir 







?^*f 



fi *^ \ ♦ . «p <* ;^i >^ 



f>^::'«-- 




:5:^^ iv 



^ i 



.^, 



■V,i 

















Book 

Copyright N^_ 



COPYRIGHT DEPOSIT. 



THE ELEMENTS 

OF THE 

MECHANICS OF MATERIALS 

AND OF 

POWER TRANSMISSION 



WILLIAM R. KING, U. S. N., Retired 

PRINCIPAL, BALTIMORE POLYTECHNIC INSTITUTE 



FIRST EDITION 

FIRST THOUSAND 



NEW YORK 

JOHN WILEY & SONS 

43 AND 45 East 19TH St. 

London: CHAPMAN & HALL, Limited 

1911 



\ 






Copyright, 191 i 

BY 

WILLIAM R. KING 



rO 



Stanbopc ipress 

F. H. GILSON COMPANY 
BOSTON. U.S.A. 




(g^CI.A287087 



PREFACE 

This book is the result of an experience of some years in class- 
room work with engineering students, and is intended for use in 
technical schools and colleges. 

It has frequently occurred to me that there is needless ob- 
scurity of statement in the average engineering text, with the 
consequent discouragement and retardation of young students. 
It has been my aim, therefore, to characterize the demonstra- 
tions in this text by completeness and simplicity of statement, 
in the belief that such treatment will greatly facilitate the study 
of more advanced works. 

The Calculus has been introduced necessarily, but only in its 
elementary form, and chiefly in demonstrations and solutions. 
Such use of it will be, I believe, beneficial to the young student 
in showing him possibilities in the application of the subject to 
practical problems. 

The book is in two parts. Part I is devoted to the elements 
of the Mechanics of Materials, and Part II to the elements of 
Power Transmission. It has been the aim to present both 
subjects only to the extent that will impart such a working 
knowledge of the fundamentals as will enable the student to 
grasp more extended works without aid. 

Much of the matter on iron and steel in Chapter IX, Part I, 
has been taken by permission from Durand's '' Practical Marine 
Engineering," and the chief sources of reference have been the 
works of Goodman and Unwin. 

I am under obligation to my assistants in engineering, WilHam 
L. De Baufre, Charles E. Conway, and Samuel P. Piatt, — to 
the first two for valuable aid and suggestion and to the third 
for the care with which he made the tracings for the cuts. 

WILLIAM R. KING. 

Baltimore, July 4, 1911. 



CONTENTS 

PART I 

Chap. Page 
I. Moments. Center of Gravity. Moment of Inertia. Radius 

OF Gyration i 

II. Bending Moment. Bending-moment Diagram. Shear. Shear 

Diagram 25 

III. The Theory of Beams. Beam Design 62 

IV. The Deflection of Beams 73 

V. Columns. Shafts 89 

VI. Internal Work Due to Deformation. Suddenly Applied 

Loads 105 

"VTI. Graphic Statics: System of Lettering. Force Diagram. Fu- 
nicular Polygon 114 

VIII. Framed Structures. Reciprocal Diagram 145 

IX. Engineering Materials 184 

X. Testing Materials 207 

'part II 

I. Transmission of Power by Belts and Ropes 219 

11. Transmission by Toothed Wheels 241 

Index 257 



PART I 

THE ELEMENTS OF THE MECHANICS 
OF MATERIALS 



CHAPTER I 

MOMENTS. CENTER OF GRAVITY. MOMENT OF 
INERTIA. RADIUS OF GYRATION 

1. Introductory. — The mechanics of materials, embracing 
the strength of materials, is an all-important subject to the 
engineering student. It includes all the calculations connected 
with the design' of machines which admit of motion between 
some of their parts in the transmission of force, thus involv- 
ing dynamical principles ; and of the design of structures 
which remain in the static state of rest. It presupposes 
for the student a course in mechanics, but the questions 
of moments, center of giavity, moment of inertia, and radius 
of gyration are of such frequent application in mechanical 
design that a partial review of those subjects is given in this 
chapter. 

2. Moments. — The moment of a force acting on a body 
may be defined as the tendency of the force to turn the body 
about a point, or about a fixed axis, and its measure is the 
product of the force by the perpendicular distance from the 
point, or from the axis, to the line of action of the force. 
The point, or axis, about which the moments are taken is 
called the center of moments, and the perpendicular distance 



2 THE ELEMENTS OF MECHANICS OF MATERIALS 

from the line of action of the force to the center of moments 
is called the arm of the force. 

If there be a number of forces acting on the body, those 
tending to turn it in one direction may be regarded as positive 
and those tending to turn it in the opposite direction as nega- 
tive. By common consent forces with a turning tendency in a 
clockmse direction are termed positive and those with a contra- 
clockwise tendency are termed negative. It is immaterial which 
kind of force is termed positive and which negative, but having 
chosen one kind as positive in any investigation the choice 
must be adhered to and the opposite kind must be regarded 
as negative. 

In Fig. I let the forces P, Q? and R act on the body in the 

directions indicated. If the body remains in equilibrium the 

^ \ underlying principle of moments asserts that the 

o^--- v---- algebraic sum of the moments of the forces about 

\ \ ^' ciny point as a center, or about any line as an 

\ V \ axis, must be zero, the point and the line being 

^-""^ ^i in the same plane. In other words, the sum of 

^ the clockwise moments must equal the sum of 

the con traclock wise moments. 

Let moments be taken about the point 0. The force Q tends 

to turn the body in a clockwise direction about and will be 

regarded as positive, and the tendencies of the forces P and R 

are to turn it in a con traclock wise direction about and will 

be regarded as negative. The equation of moments will then 

be 

Qq - Pp - Rr = o. 

This follows directly from the meaning of the word equilibriumy 
which implies that the body is at rest, and that condition can 
only result when there is no tendency to turn the body about 
the point 0; that is, when the algebraic sum of the moments 
about is zero. Should the line of action of a force pass 



CENTER OF GRAVITY 3 

through the center of moments, the moment of that force would 
vanish. 

In expressing the value of moments, the units of force, mass, 
area, and volume are placed first and the length units afterward. 
For example, a moment may be expressed as so many pounds- 
feet, and thus avoid confusion with work units. 

3. Center of Gravity. — The center of gravity of a body or 
of a system of bodies is a point on which the body or system 
will balance in all positions, supposing the point to be sup- 
ported, the body or system to be acted on only by gravity, and 
the parts of the body or system to be rigidly connected to the 
point. 

It follows from this definition that all the particles of a body 
or of a system of bodies are acted on by a system of parallel 
forces, gravity acting on each, and that the algebraic sum of 
the moments of these forces about a line must be zero when the 
line passes through the center of gravity of the body or system; 
otherwise the body or system would not balance. 

Let two heavy weights P and Q be situated as shown in Fig. 2. 

Join them with a straight Hne and divide it at C so that — = ? • 

Q P 
If the weights be joined by a rigid rod without weight, the 
system will, by the principle of the 
lever, balance when supported at C. 
The center of gravity of the system is 
therefore at C As the resultant of 
the weights is P -\- Q, the pressure 
on the support will h^ P -\- Q. 

The center of gravity of a uniform straight rod is evidently 
at the middle point of its length, and when considering a body 
at rest we may assume its whole mass to be concentrated at 
its center of gravity. 

Let the weights P and Q, Fig. 3, be attached to a balanced 




4 THE ELEMENTS OF MECHANICS OF MATERIALS 

rod as shown. Then, as we have just seen, the center of 

P n 
gravity will be so situated that — = — , or Pm = On; that is, 

Q m 

the algebraic sum of the moments is zero. 

The distance of the center of gravity of a system from a 
point or from a line may readily be determined by moments. 
Thus, in Fig. 3, let ^ denote the distance of the center of grav- 
ity of the system from O, and let p and q be the distances, 




C. of G. 



<!> 



Fig- 3- 



respectively, of the centers of gravity of P and Q from 0. 
We have from the figure, m = x — p and n = q — x. Since 
the center of gravity must be between P and Q, and since the 
turning tendency of P about the center of gravity is contra- 
clockwise and that of Q clockwise, we shall have by the prin- 
ciple of moments 

P(x-p)=Q{q-x), 

whence 

Pp±Qq, 

P+Q 

If the system be extended the result 

_ _ Pp+Qq + Rr + etc. 
^ P + e + i^ + etc. ' 

will be obtained. This formula is extensively used in the 
solution of problems. 

If a sheet of uniform thickness weighing M pounds per unit of 
area be considered, the weight of any area a will be Ma pounds. 



CENTER OF GRAVITY 



and we may substitute Mai for P and Ma2 for Q and obtain 

_ ^ Maip + Ma2q ^ aip + a^g ^ aip -\- a^g 
Mai + Ma^ ai + a^ A 

in which A is the whole area. This may be expressed as follows: 
The distance, x, from O to the center of gravity is 

Sum of the moments of all the elemental surfaces about O 

^ = . - . 

Whole surface 
_ Moment of the whole surface about 
Whole surface 
Similarly, 

Moment of the whole volume about O 

X = • • 

Whole volume 

The moment of the whole surface and of the whole volume is 
obtained by integration. 

4. The center of gravity of a portion of the perimeter of a 
regular polygon or of an arc of a circle, considered as a thin wire, 
may be obtained thus: 

In Fig. 4, let L = the sum of the 
lengths of the sides of the portion of the 
perimeter taken, or the length of the arc 
in the case of a circle; R — the radius 
of the inscribed circle, or 'the radius of 
the circle; C = the chord of the arc of 
polygon or circle; F= the distance of the center of gravity 
from the center of the circle. Then, in Fig. 5, let a denote the 




Fig. 4. 




6 THE ELEMENTS OF MECHANICS OF MATERIALS 

length of a side of the polygon. The center of gravity of each 
side will be at its middle point, and distant 3^1, ^2, ys, etc., from 
the diameter of the inscribed circle. Let Xi, x^, X3, etc., denote 
the projected lengths of the sides on the diameter. From the 
similar triangles Obc and edf we have 

de : Ob = df : be, or a : R = Xii yi^ 

, Rxi 

whence yi = 

a 

Roo> 

Similarly, 3^2 = ' and so on. 



If w denotes the weight of each of the n sides of the polygon,, 
we shall have 

-r^ Pp -{- Qq -\- etc. wyi + ^^2 + etc. ^1 + ^2 + etc. 



j: — 


P + (2 + etc. 


nw n 


Substituting the values of yi, y2, ys, etc., we have 




Y = 


= — (:i:i + X2 + etc.). 
na 


But 


na = L, 


and Xi + X2 + etc. = C. 


Hence 







5. The center of gravity of a plane figure may be obtained 

graphically as follows: 

Let abcde, Fig. 6, be any plane figure. Draw eb and ec. Let 
Ai and gi, A^ and g2, and A^ and ^3 denote the areas and centers 
of gravity of the triangles eab, ebc, and ecd respectively. Join gi 
and g2. The center of gravity of the figure abce must lie in this^ 
line. From gi lay off in any direction and to any scale the dis- 
tance git equal to A2, and in the opposite direction and to the 
same scale lay off from g2 the distance g2S equal to Ai and paral- 
lel to git. Join st; then its intersection, k, with gig2 is the center 



CENTER OF GRAVITY 



of gravity of the figure ahce. Join k and g^. 
gravity of ^1 + ^2 + ^3 will lie in this line, 
in any direction a distance 
kr equal to the area Az, and 
from ^3 lay off gzv parallel 
to kr and equal to ^1 + A2. 
Join vr, and its intersection 
with kgz gives G as the cen- 
ter of gravity of the whole 
figure. 

Should the surface con- 
tain a hole, 3iS fmn in Fig. 7, 
we would proceed thus: 

Denote by Ai and gi 
and A2 and g2 the areas 
and centers of gravity of 



The center of 
From k lay off 




Fig. 6. 



the whole figure abcde and of fmn respectively. Draw gig2. 

b 




From g2 lay off to scale, in any direction, g2S equal to A\, and 
parallel to it, and on the same side of gig2, lay off gir. Join 



8 



THE ELEMENTS OF MECHANICS OF MATERIALS 



sr and extend it to meet gig2 produced at G. Then the point G 

is the required center of gravity. 

The truth of these graphic methods may be shown as follows : 
Let Oi and O2, Fig. 8, be the positions of the centers of gravity 

of two areas Ai and A2 respectively, and let their common 
d 




- ^ Fig. 8. 

center of gravity be situated at g, distant ri from Oi and ^2 
from O2. By the principle of moments we shall have Airi = 
^2^2. In any direction from O2 lay off O^b, whose length repre- 
sents the area Ai to some selected scale, and from Oi lay ojff 
Old parallel to O2& and of such length as to represent the area A2 
to the same scale. The triangles Oidg and 02bg are similar, and 

we have 

O26 : Old = 02g : Oig, or Ai : A2 = r2 : n, 

whence Aifi = ^2^2. 

That is, the line joining b and d passes through g. It will be 

observed that the lines O2& and Oid are laid off on opposite sides 

of the line joining Oi and O2 and 
at opposite ends of their respec- 
tive areas and at any convenient 
angle. Should one of the areas, 
as A2, be negative, i.e., the area 
of a hole, or of a part cut out 

of the surface, then O2& and Oid must be laid off on the same 

side of O1O2, as shown in Fig. 9. 




CENTER OF GRAVITY 



6. Applications. — To show the application of the foregoing 
principles to finding centers of gravity, the solutions of a few 
problems will be given. 

Example I. — Find the center of 
gravity of a triangle. 

Conceive the triangle, Fig. lo, to 
be divided into a great number of 
very narrow strips drawn parallel to 
one of the sides, as B. The center 
of gravity of each strip will be at its 
middle point, therefore the center of gravity of the triangle 
will lie in the locus of these middle points; that is, in the 
median Oc. 

The elemental area of the triangle = h 'dh. 

The moment of the elemental area = h'b » dh. 

From similar triangles we have 




Fig. lo. 



b:B = h: H 



whence, b = f • 



Then 



Moment of elemental area 



B'h^'dh 
H 



B r^ 

Moment of the whole ^ea = ^ / h^dh = 



BE"- 



The area of the triangle 



E 
BE 



The distance, x^ of the center of gravity from the apex is 

BE'' 

- _ Moment of whole area _ 3 _ 2_E 
Whole area BE 3 



That is, the center of gravity is at a perpendicular distance 
below the apex equal to two-thirds of the altitude, and must 



lO 



THE ELEMENTS OF MECHANICS OF MATERIALS 



therefore be at g, the intersection of the median, Oc, and the 

parallel to the base distant — from the apex. From similar 

3 
triangles it is seen that Og is two- thirds the length of the median, 

so that the center of gravity is on the median at two-thirds its 

length from the apex. 

Example II. — To find the center of gravity of a portion of 

a regular polygon or of a sector of a circle when considered as 

a lamina or thin sheet. 




Fig. II. 

Referring to Fig. ii, and using the same notation as in Art. 4, 

2 R 
we shall have as the distance of the center of gravity of 

. ^ 
each of the triangles from the center of the inscribed circle, and 

they will be distant yi, y2, ys, etc., from the diameter. The base 

of each of the triangles is a, and the projected lengths of these 

bases on the horizontal diameter are Xi, 0C2, X3, etc. From the 

similar triangles Obc and edf we have 



similarly. 



2R , 2 Rxi 
: a = yi : Xi, whence yi = ; 

3 3^ 

2 RX2 J 

72 = ' and so on. 

3^ 



If w denotes the weight of each of the n triangles of the polygon 
we shall have 

nw n 2)^a 3 L 



CENTER OF GRAVITY 



II 



Example III. — A square is divided into four equal triangles 
by diagonals intersecting at 0; if one triangle be removed, find 
the center of gravity of the figure formed 
by the three remaining triangles. 

Let w denote the weight of each of the 
remaining triangles, and let a denote the 
side of the square, Fig. 12. The weight 
2woi the side triangles may be supposed 
concentrated at their common center of 
gravity 0. The distance of the center of 
gravity of the lower triangle from is 




Fig. 12. 



- X - = - ' by Example I. 
2 3 3 



Then if x denotes the distance of the 



center of gravity of the three remaining triangles from 0, we shall 
have 



X = 



Pp + Qq 



2W Xo ^wX 



3 = ^. 

9 



P -\-Q 2W -\-w 

That is, the required center of gravity is distant one-ninth the 

side of the square from the center of the square. 

Example IV. — A quarter of the area of a triangle is cut off 

by a line drawn parallel to the base. Find the center of gravity 

of the remaining quadrilateral. 

Let EF be parallel to the base BC, 
Fig. 13, and let the triangle AEF be 
the part cut off. The required cen- 
ter of gravity will lie in the median 
AD. The triangles AEF and ABC 
are similar, and are to each other 
in area as i 14. Since the areas of 

similar triangles are to each other as the squares of homologous 

sides, we have 

AD 
2 




Fig. 13. 



AG"" •'AD' = ^ 



4, whence AG = 



12 



THE ELEMENTS OF MECHANICS OF MATERIALS 



Let X denote the distance of the required center of gravity of 
the quadrilateral BEFC from A. Then 

^^ 2XAD _ AD^2 

-_ PP-Qq ^^—J— '^ 2 ^, _ AD{S-i) 

P-Q 4-1 9 



==1ad. 

9 



4 - I 

That is, the required center of gravity is in the median AD and 
at seven-ninths of its length from A . 

Example V. — Find the center of gravity of a cone. 
Conceive the cone of Fig. 14 to be made up of a great number 
of thin sections, each parallel to the base. The center of gravity 

of each of these sections will be 
at its center, therefore the center 
of gravity of the cone will lie in 
the locus of the centers of these 
sections; that is, it will lie in 
the line joining the vertex with 
the center of gravity of the 
base. 

The elemental volimie of the 

cone is irr^dh, and the moment of 

From similar triangles we have 




Fig. 14. 
the elemental volume is irr^hdh. 

R:r=H:L 



Rh 
whence r = — 
H 



Then by substitution we have 



Moment of elemental volume = 



Moment of whole volume 
Volume of cone = 






irR^hHh 

H 

hHh 



rRm' 



If X denotes the distance of the center of gravity of the cone 
from the vertex, we shall have 



CENTER OF GRAVITY 13 

_ _ Moment of whole volume _ 4 _ 3^ 
Whole volume ttR^H a 

3 
That IS, the center of gravity is at a perpendicular distance 
below the apex equal to three-fourths of the altitude, and must 
therefore be at g, the intersection of the line Oc joining the apex 
with the center of gravity of the base and the parallel to the base 

distant - — from the apex. From similar triangles it is seen 

that Og = ^ — , so that the center of gravity of the cone is in 

4 

the line joining the vertex with the center of the base and at 
three-fourths its length from the vertex. 

Example VI. — Find the center of gravity of a semicircular 
arc, or wire. 




Fig. IS. 



7?r* 
In Fig. 15 we have, from Art. 4, 7 = — - , in which Y denotes 

the distance of the center of gravity from the center, L the 
length of the arc, and C the chord of the arc. 
Then 

y^ RX2R _2R_D 
ttR it tt 



14 THE ELEMENTS OF MECHANICS OF MATERIALS 

PROBLEMS 

I. A rod 3 feet long and weighing 4 pounds has a weight of 2 pounds 
placed at one end; find the center of gravity of the system. 

^, Ans. One foot from weighted end. 

" . •2. Find the center of gravity of a uniform circular disc out of which 
another circular disc has been cut, the latter being described on a radius of 
^the former as a diameter. 

Ans. One-sixth of radius of large circle from center. 

3. A heavy bar 14 feet long is bent into a right angle so that the lengths 
of the portions which meet at the angle are 8 feet and 6 feet respectively; 
show that the distance of the center of gravity of the bar so bent from the 
point of the bar which was the center of gravity when the bar was straight, 

is feet. 

7 

4. The middle points of two adjacent sides of a square are joined and 
the triangle formed by this straight Hne and the edges is cut off; find the 
center of gravity of the remainder of the square. 

Ans. -ii of diagonal of square from center of square. 

5. A piece of uniform wire is bent into the shape of an isosceles tri- 
angle; each of the equal sides is 5 feet long, and the other side is 8 feet long; 
find the center of gravity. Ans. j\ of the altitude from the base. 

6. Find the center of gravity of a figure consisting of an equilateral 
triangle and a square, the base of the triangle coinciding with one of the 
sides of the square. 

Ans. At a distance from the base of the triangle equal to ^—^ — 

26 
times the base. 

7. A table whose top is in the form of a right-angled isosceles triangle, 
the equal sides of which are 3 feet in length, is supported by three vertical 
legs placed at the corners; a weight of 20 pounds is placed on the table at a 
point distant 15 inches from each of the equal sides; find the resultant 
pressure on each leg. Ans. 8J, 8^, 3^ lbs. 

8. ABCD is a quadrilateral figure such that the sides AB, AD, and the 
diagonal AC are equal, and also sides CB and CD are equal; find its center 
of gravity. 

Ans. — units from C, in which a = AB and h = CB. 

6a 

9. ABC represents a triangular board weighing 10 pounds. Suppose 
weights of 5 pounds, 5 pounds, and 10 pounds are placed at ^, B, and C 
respectively. Where is the center of gravity of the whole ? 

Ans. At five-ninths of the median drawn from C. 



CENTER OF GRAVITY 1 5 

10. A rod of uniform thickness is made up of equal lengths of three 
substances, the densities of which taken in order are in the proportion of 
1,2, and 3 ; find the position of the center of gravity of the rod. 

Ans. At seven-eighteenths of the length of the rod from the end of the 
densest part. 

11. Find the position of the center of gravity of a piece of wire bent to 
form three-fourths of the circumference of the circle of radius R. 

Ans. On a line drawn from the center of the circle to a point bisecting 
the arc, and at a distance 0.3 R from the center. 

12. A thin wire forms an arc of a circle the radius of which is 10 inches, 
and subtends an angle of 60°; find the distance of the center of gravity 

from the center. Ans. — ins. 

IT 

13. Find the position of the center of gravity of a balanced weight 
having the form of a circular sector of radius R, subtending an angle of 
90°. Ans. 0.6 R from center of circle. 

14. Find the center of gravity of a semicircular lamina, or sheet, of 

2 D 
diameter D. Ans. — from the center. 

15. A square stands on a horizontal plane; if equal portions be removed 
from two opposite corners by straight lines parallel to a diagonal, find the 
least portion which can be left so as not to topple over. 

Ans. Three-quarters of the area of the square. 

16. Find the center of gravity of a trapezoid. 

Ans. On the line joining the middle points of the bases, and at a per- 
pendicular distance from the upper base equal to , and from the 

lower base, — , B and b being the lower and upper bases respec- 

3 B + b 

tively, and H the altitude. If the distance be measured on the line 6* joining 

S* 77 

the middle points of the bases, then - must be substituted for — • 

3 3 

17. Find the center of gravity of a pyramid. 

Ans. In the line joining the vertex with the center of gravity of the 
base and at three-fourths its length from the vertex. 

18. Find the center of gravity of a frustum of a cone. 

Ans. In the line joining the centers of gravity of the upper and lower 

bases, at a distance from the upper base equal to — • ^-—z r- ; 

4. R^ + Rr + r^ 

and from the lower base — • "^-—r — ^^^— , R and r being the "radii of 

4 R^ -\- Rr + r 



l6 THE ELEMENTS OF MECHANICS OF MATERIALS 

the lower and upper bases respectively, and H the altitude. These 
results are true for the frustum of any pyramid by substituting B for R 
and h for r, B and h being homologous sides of the lower and upper bases 
respectively. 

19. A lever safety valve is required to blow off at 70 pounds pressure 
per square inch. Diameter of valve, 3 inches; weight of valve, 3 pounds; 
short arm of lever, 2.5 inches; weight of lever, 11 pounds; distance of 
center of gravity of lever from fulcrum, 15 inches. Find the distance at 
which a cast-iron ball 6 inches in diameter must be placed from the ful- 
crum. The weight of the ball-hook is 0.6 pound, and that of a cubic inch 
of cast iron 0.26 lb. Ans. 35.48 ins. 

20. A steel safety-valve lever i inch thick and 50 inches long tapers 
from 3 inches in depth at the fulcrum to i inch at the end. It overhangs 
the fulcrum 4 inches, the overhang having no taper. Diameter of valve, 
4 inches; weight of valve, 4.75 pounds; short arm of lever, 3.5 inches. 
Find the distance from the fulcrum at which a cast-iron ball 9.5 inches in 
diameter must be placed in order that steam shall blow off at 125 pounds 
pressure per square inch. Weight of the ball-hook, 1.3 pounds; weight of 
a cubic inch of steel, 0.28 pound; weight of a cubic inch of cast iron, 0.26 
pound. Ans. 42.35 ins. 

21. A trapezoidal wall has a vertical back and a sloping front face; 
width of base, 10 feet; width of top, 7 feet; height, 30 feet. What hori- 
zontal force must be applied at a point 20 feet from the top in order to 
overturn it, i.e., to make it pivot about the toe? Width of wall, i foot; 
weight of masonry in wall, 130 pounds per cubic foot. 

Ans. 18,900 lbs. 

22. Find the height of the center of gravity from the base of a column 
4 feet square and 40 feet high, resting on a tapered base forming a frustum 
of a square-based pyramid 10 feet high and 8 feet square at the base. 

Ans. 20.4 feet from base 

Problems 22,, 24, and 25 are to be solved graphically. 

23. Find the position of the center of gravity of an unequally flanged 
section; top flange, 3 inches wide, 1.5 inches thick; bottom flange, 15 
inches wide, 1.75 inches thick; web, 1.5 inches thick; total height, 18 inches. 

Ans. 5.72 inches from the bottom edge. 

24. Find the height of the center of gravity of a T-section from the 
foot, the top crosspiece being 12 inches wide and 4 inches deep; the stem, 
3 feet deep and 3 inches wide. 

Ans. 24.16 inches. (Check the result by seeing if the moments are 
equal about a line passing through the section at the height found.) 



MOMENT OF INERTIA 1 7 

25. A square board weighs 4 pounds, and a weight of 2 pounds is placed 
at one of the corners. Find the position of the center of gravity of the 
board and weight. 

Ans. On the diagonal drawn from the weighted corner and at two- 
thirds its length from the opposite corner. 

7. Moment of Inertia. — The moment of inertia of a surface 
is the sum of the products of each elemental area of the sur- 
face by the square of its distance from an axis about which the 
surface is supposed to be revolving. If, instead of a surface, 
we have a body, then we must substitute the elemental volume 
for the elemental area in finding the moment of inertia. The 
moment of inertia varies according to the position of the axis, 
being smallest when the axis passes through the center of 
gravity. 

It would be a crude and inaccurate method of finding the 
moment of inertia by actually dividing an area or a volume 
into its elements, multiplying each by the square of its distance 
from the axis, and then finally taking the sum of these products. 
We can, however, find the moment of inertia with accuracy by 
means of integration. 

The strength of a beam or of a column depends upon the form 
as well as the area of its section, and in all calculations respecting 
the strength of beams and columns, the factor which gives 
expression to the effect of the form of the section is its moment 
of inertia about an axis passing through the center of gravity. 

The moment of inertia of a surface is universally denoted by 
/, the axis being in the plane of the surface and passing through 
its center of gravity. When the axis is perpendicular to the 
plane of the surface, the moment of inertia is then termed the 
polar moment of inertia, and is denoted by I p. 

If we denote the product of a force, area, volume, or weight by 
its arm as the first moment, or simply the moment, of the force, 
area, volume, or weight, then we may conveniently denote the 



i8 



THE ELEMENTS OF MECHANICS OF MATERIALS 



product of the force, area, volume, or weight by the square of 

its arm as the second moment of the force, area, volume, or weight. 

Thus the moment of inertia is sometimes known as the second 

moment. 

8. Relation between Moments of Inertia about Parallel 

Axes. — There is a relation between moments of inertia of a 

surface about parallel axes that is useful in the solution of 

problems. 

Let / denote the moment of inertia of any surface about an 

axis through the center of gravity 
of the surface, I' the moment of 
inertia of the same surface about 
any other parallel axis, A the area 
of the surface, and D the per- 
pendicular distance between the 
axes. Then we shall have /' = / + 
AD\ 

For, let yy, Fig. i6, be the axis 
through the center of gravity, and 
y'y the axis parallel to yy, both 
being in the plane of the surface. 
Denote the elemental areas of the surface by ai, a^, a^, etc., 

and their distances from the axis yy by ri, f2, r^, etc. 

Then 

r = ai(ri - DY + a^ (D - nf + ^3(2) + nY + etc. 

+ as (Z)2 + 2 rgZ) + rs') + etc. 

= {aiTi^ + (hr2^ + a^n^) + (ai + 02 + as)D^ 
-2D {air I + 02^-2 - a^rs). 

The first term of the second member of this equation is, by 
our definition, the moment of inertia, /, of the surface about 
the axis yy through the center of gravity; the second term 




Fig. 16. 



RADIUS OF GYRATION 



19 




Fig. 17. 



becomes AD^, in which A denotes the whole area; and the third 
term will reduce to zero, because the quantity within the paren- 
thesis is the algebraic sum of the moments of the elemental 
areas about a line passing through their center of gravity. 
Therefore I' = I -\- AD\ 

9. Suppose an axis perpendicular to the plane of the surface 
represented in Fig. 17 to pass through 0. Let a be an elemental 
area of the surface, distant r from 0, 
and let X and Y be rectangular axes 
passing through and lying in the 
plane of the surface. Then the polar 
moment of a is /^ = ar"^. The moment 
of inertia of a about X is Z^. = ay^, 
and similarly ly = ax^. But we have 
^2 _j_ ^2 ^ r^, therefore ax^ + ay^ = ar'^. 
That is, the polar moment of inertia 
of any surface is equal to the sum of 
the moments of inertia of the surface about any two rectangular 
axes lying in the plane of the surface and passing through the 
polar axis of revolution; that is, 

10. Radius of Gyration. -^ We have seen, Art. 8, that I = 
diri^ + (hr2^ + ^3^3^ + etc., in which ai -\- (h + (I3 + etc. = A, 
the whole area. If we can conceive the whole area to be con- 
densed into a single particle, distant K from the axis of rotation, 
we shall have / = AK^. This imaginary point at which the 
particle is supposed to be situated is known as the center of 
gyration, and its distance, K, from the axis is the radius of 

gyration. 

volume, V, or the weight, W, may be substituted for the area, A. 

11. Illustrations. — To illustrate the methods of finding mo- 
ments of inertia and radii of gyration a few examples follow. 



We have then, i^ = y — ? in which the mass, M, the 



20 



THE ELEMENTS OF MECHANICS OF MATERIALS 



Example I. — To find the least moment of inertia and least 
radius of gyration of the surface of a parallelogram. 

The least moment of inertia is that about an axis passing 
through the center of gravity and parallel to the bases. Let X X, 
Fig. 1 8, be the axis through the center of gravity and parallel 
to the base B. It bisects the altitude h. 




Fig. 1 8. 



Suppose the surface to be divided into an infinite number of 
strips parallel to the axis, each of thickness dy. Consider one 
of these strips distant y from the axis. 

Then Elemental area = Bdy, 



Second moment of elemental area = dl = By'^dy. 



Then 



h 
>3n2 



'-f>'i%-ii-i-B]-^ 



If the axis YY parallel to the base b be taken, we shall have 



/ = 



12 



RADIUS OF GYRATION 21 

For the radius of gyration we shall have 



iS: = v/4 = V/-^ = f v^ for the axis XX, 



and 



K = v/-^. = T ^3 for the axis YY. 
^ 12 bH 6 



If the moment of inertia with respect to an axis coinciding 
with the base be desired, we have /' = / + AD^j in which D 
denotes the perpendicular distance between the axes, denoted in 

this instance by - when B is the axis, and by — when b is the axis. 



Then 

and 
Also 
and 



I' = [- Bh X — = for the base B as axis, 

12 4 3 

/ = \-oH X — = for the base b as axis. 



12 4 3 

"as axis, 



K' = V -^ = - V3 for the base B 
^ sBh 3 

K' = V— tt; = ~ "^3 for the base b as axis. 
^3^3 3 



Example II. — Find the moment of inertia and radius of gyra- 
tion of a circular surface about an axis 
passing through the center and perpen- 
dicular to the plane of the surface. 

The axis being perpendicular to the sur- 
face it is the polar moment of inertia that 
is required. 

Conceive the circular surface of Fig. 19 
to be made up of an infinite number of 
concentric circular strips, each of thick- ^^' ^^ 

ness dr. Consider the strip distant r from the center. 

Then Elemental area = 2 irrdr, 

Second moment of elemental area = dl = 2 irr^rdr. 




22 



Then 



THE ELEMENTS OF MECHANICS OF MATERIALS 

»i2 



f^dr 



2 



7rD\ 

32 



K= Jl = J ^^' = ^^^ = i^V2 . 

V^ VsttZ^^ 42 

Example III. — Find the polar moment of inertia and radius 
of gyration of a right circular cone about its axis. 

Conceive the volume of the cone, 
Fig. 20, to be made up of an infinite 
number of circular strips, each of 
thickness dh. Consider the strip of 
radius r and distant h from the 
vertex. 
Then 

Elemental volume = irr^dh. 
^ig- 20. If we imagine the elemental volume 

to be condensed into a single particle, the distance of this 
particle from the axis will be the radius of gyration K, which 

for the circle is - — - . 




Hence, 
Second moment of elemental volume = dlp= irr^KHh = 



irr^dh 



Then 



2t/n 



r^dh. 



By similar triangles, - 
Then 



whence h = -—j and dh 
K 



Edr 
R 



2R^o 10 



" 2RJo 

K = \/|'= 




RADIUS OF GYRATION 



23 



PROBLEMS 

Find the moment of inertia and radius of gyration of the following: 
I. Triangle about an axis through vertex and parallel to base. 

Ans. /'=^; K = ^^'2. 

4. 2 



2. Triangle about an axis through the center of gravity and parallel 
to base. 

3. Triangle about an axis coinciding with base. 



Ans. 7 = —^; Z = — : v 18. 

36 15 



Ans. r = ; A = — v6. 

12 o 



4. Trapezoid about an axis coinciding with small base. 



Ans.r^^S^ 



B-+.b) . _ / 3B + 



5. Trapezoid about an axis coinciding with large base. 



...r = -(iA±^;.=.v/^ 



6. Trapezoid about an axis through center of gravity and parallel to 



base. 



Ans I - B'iB^ + 4Bb+b^) . _ 
7. Square about its diagonal. 



8. Circle about a diameter. 



9. Hollow circle about a diameter. 



6{B + b) 



V2{B^+4Bb + b^). 



12 



Ans. I =-; K =- V12. 
12 



Ans. I =rDl, K ^^ 

64 4 



Ans. I ^~{D^- d^) ; K = 
64 



V Z)2 + J2 



Find the polar moment of inertia and radius of gyration of the fol- 
lowing named surfaces: 

10. Parallelogram about a pole passing through its center of gravity. 

BE ,^„ . „„, ^^ I h^+B^ 



Ans. Ip ==^(^2 + ^2); K 



12 



II. Hollow circle about a pole passing through center. 



Ans. /p =-^(Z)4_^4); 
2>2 



K = SI^ 



2+f^2 



24 THE ELEMENTS OF MECHANICS OF MATERIALS 

Find the polar moment of inertia and radius of gyration of the follow- 
ing named solids: 

12. Cylinder about a pole coinciding with its axis. 

2>^ 4 

13. Hollow cylinder about a' pole coinciding with its axis. 

32 






14. Sphere about a diameter 



^... /,=^^;i^=^^- 



IS 5 

15. A bar of rectangular section about a pole passing through the 
center of figure. 

12 ^ V 12 



CHAPTER II 

BENDING MOMENT. BENDING MOMENT DIAGRAM. 
SHEAR. SHEAR DIAGRAM. 

12. Bending Moments. — When forces act on a body in such 
a manner as to tend to give it a spin or a rotation about an axis 
without any tendency to shift its center of gravity, the body is 
said to be acted on by a couple. A couple consists of two parallel 
forces of equal magnitude acting in opposite directions but not 
in the same line, the arm of the couple being the perpendicular 
distance between the lines of action of the forces. 

A beam is subjected to a bending moment when it is so acted 
upon at its ends by equal and opposite couples that there is a 
tendency to turn it in opposite directions. 




w 



R + W 



Fig. 21. 

Thus, the beam mn of Fig. 21 is acted on by the equal and 
opposite couples, R and R, and W and W, the tendency being 
to turn the beam in opposite directions about the point r; that 
is, to bend it at r. In Fig. 22, the couples whose moments are 
Ri X mr and R2 X nr have the same effect on the beam as those 
of Fig. 21. The beam of Fig. 21 is called a cantilever, from the 

25 



26 THE ELEMENTS OF MECHANICS OF MATERIALS 

nature of its support, while the beam of Fig. 22 is called a simple 
beam. The forces R, Ri and R2 are the support reactions. 



w 



R1+R2 

0) 



Fig. 22. 

13. General Case of Bending Moments. — The 

moment at any section of a beam is the algebraic sum of the moments 
of all the external forces acting to the left of the section. It is 
assumed that forces acting upward are positive, and those act- 
ing downward are negative, so that bending moments may be 
positive or negative. 

In graphical constructions the signs of bending moments are 
of the first importance and are determined by the following 
rule : 

Bending moments which tend to bend a beam or cantilever 
concave upward, -^^ — ^, are regarded as positive, and when they 
tend to bend in the reverse way, ^ — ^, they are negative. 

It should be observed that it is merely to avoid confusion in 
the construction of diagrams that the external forces to the left 
of the section were considered when defining the bending moment 
at any section of a beam. Moments may equally well be taken 
to the right of the section and the same value be obtained for 
the bending moment at the section. When bending moments 
are obtained by calculation rather than by construction, the 
side involving the least calculation in taking moments should 
be selected, though the calculations for both sides afford a posi- 
tive check as to the accuracy of the work. To avoid confusion 
with work units, bending moments are expressed in pounds- 
inches, pounds-feet, or tons-inches, as may be found most con- 
venient. 



BENDING MOMENTS 



27 



The beam of Fig. 23 is supposed to be without weight. The 
bending moment at section E is, taking moments to the left 
of £, 

M = RiXaE-WiXqE-W2X pE, (i) 

or, taking moments to the right of E, 

M = R,XbE-WzX rE. (2) 

Suppose the distances and weights to be as shown in the 
figure. Before finding the bending moments we must first find 
the reactions, Ri and R2. 



lbs. 



Wo^SO lbs. 
-1.5^ 



Q Q 



W3=801bs. 

—2^ — 



m 



ik6 



Fig. 23. 

Taking moments about the left support, we have 
R2 X 10.5 = 80 X 8.5 + 50 X 5 + 60 X 2, 
whence R2 = 100 pounds. 

Taking moments about the right support, we have 
Ri X 10.5 = 60 X 8.5 + 50 X 5-5 + 80 X 2, 
whence Ri = 90 pounds, 

which might have been expected, since Ri + R2 should equal 
Wi + W2-\- Ws. 

By substitution in equation (i), we have for the bending 
moment at E, 

M = go X 6.S — 60 X 4-5 — 50 X 1.5 = 240 pounds-feet. 
Substituting in equation (2), 

M = 100 X 4 — 80 X 2 = 240 pounds-feet. 
It is thus seen that the bending moment at a section is the same, 
whether the moments be taken to the left or to the right of the 



28 



THE ELEMENTS OF MECHANICS OF MATERIALS 



section. In this instance the taking of moments to the right 
involved the least calculation. 

14. Bending Moment Diagrams. — A diagram may be made 
to show graphically the bending moment at any section of a 
beam. Such diagrams are known as bending-moment diagrams. 



Wi=6on)s. 



W2=50n)S. I W3=801T)s. 
-^^^.5^-4^ 2' ^ 2- 




Fig. 24. 



For example, take the beam of Fig. 23. The bending moment 
atgis 

If = i?i X 2 = 90 X 2 = 180 Ibs.-f t. ; 

at /> M = i?i X 5 - PFi X 3 = 90 X 5 - 60 X 3 = 270 Ibs.-ft.; 
at ;- M = RiXS.s-WiX 6.5 - W2 X 3.5 

= 90 X 8.5 — 60 X 6.5 — 50 X 3.5 = 200 Ibs.-ft. 

If, on a base line a^b\ Fig. 24, and to a scale of i inch = 200 
pounds-feet, we erect ordinates to represent these bending 
moments, and then join their extremities with the broken line 
a'q'p'r'h' , the inclosed figure is a diagram whose ordinates 



SHEAR DIAGRAMS 



29 



beneath any section of the beam will measure the bending 
moment at the section to the scale adopted. Thus the ordinate 
for the bending moment at q measures ift = 0.9 inch; that at p, 
f oS = 1-35 inches; and that at r, ^%% = i inch. The ordinate be- 
neath E measures 1.2 inches, which, to scale, represents a bend- 
ing moment of 1.2 X 200 = 240 pounds-feet, as already found by 
calculation. 

15. Shear. — Shearing stresses exist when couples, acting 
hke a pair of shears, tend to cut a body between them. Beams 
acted on by couples are subjected to shearing stresses as well as 
to bending moments, the latter being far more important in 
beams of lengths ordinarily encountered. The failure of very 
short beams will be invariably from shear. 

The shear at any section of a beam or of a cantilever is equal to 
the algebraic sum of the forces to the left of the section, the upward 
forces being regarded as positive and the downward negative. 

16. Shear Diagrams. — Diagrams made to show graphically 
the amount of the shear at any section of a beam are known as 
shear diagrams. In their construction, attention must be paid 
to signs, so that in cases where the shear is partly positive 
and partly negative the positive part may be placed above the 
shear axis, or base line, and the nega- ar^w 
tive part below. 

The failure from overload of a short 
cantilever. Fig. 25, will be from shear, 
the projecting part shearing off bodily 
from the part built in. According to 
our definition the shear at all sections of 
the beam is constant and equal to — W. 
The shear is therefore represented 
graphically by the small rectangular shaded diagram of Fig. 25, 
each of the ordinates of which is equal to —W. As the shear 
diagram is negative, it is placed below the shear axis mn. 




30 



THE ELEMENTS OF MECHANICS OF MATERL\LS 



In the case of the simple beam of Fig. 26, a failure from shear 
will occasion the part between the supports to sHde down, as 
sho\\TL by the dotted lines. By our definition the shear at 
any section to the left of the load W is Ri, and at any section 
between W and the right support it is Ri — W = — R2, since 
Ri + R2 =W. 

The shaded part of Fig. 26 represents the shear, and shows 
that it changed sign under the load, the positive part being placed 
above the shear axis fmi and the negative part below. 

As in the case when defining bending moments, only the 
external forces to the left of the section were considered when 
defining the shear at the section, but this, as in the case of 
bending moments, was only for convenience. If Wg denotes 
the sum of all the loads between the left support and a given 
section, and W/ the sum of all the loads between the section 
and the right support, then evidently Ri -\- R2 = W^ + W/, 
whence Ri - W, = - {R2 - W/). The first member of this 
equation is the algebraic sum of the forces to the left of the 

given section and is the shear at 
the section, and the second member 
is the negative of the algebraic sum 
of the forces to the right of the sec- 
tion. The shear at a given section 
is, then, the algebraic sum of the ex- 
ternal forces to the right or to the 
left of the section, but with contrary 
signs. 

Let the beam of Fig. 27 be loaded 
wdth TFi, W2, W3 Sit distances di, ^, ds 
respectively from any given section 
a. Denote the support reactions by Ri and R2, and their dis- 
tances from a by ri and ^2 respectively. 

To construct the shear diagram of the beam we select a base 




Fig. 26. 



BENDING-MOMENT AND SHEAR DIAGRAiMS 



31 



line, or axis, mn, and having determined to an appropriate scale 
the values of Ri and R2 from the given weights Wi, W2, and Wsy 
we proceed as follows: 

By our definition the shear at any section to the left of Wi 
is Ri, which gives the point b of Fig. 28. Immediately the 




Fig. 29. 

point of application of Wi is passed the shear becomes Ri — Wi, 
giving the point c, and this shear continues imchanged up to 
the point of appHcation of W2, but immediately this point is 
passed the shear becomes Ri — Wi — W2, giving the point d, 



32 THE ELEMENTS OF MECHANICS OF MATERIALS 

the shear becoming negative. This shear continues unchanged 
until the point of application of W3 is passed, when the shear 
becomes Ri — Wi — W2 — W^ = — R2, giving the point e. This 
shear continues unchanged to the right support, giving the point 
and completing the diagram. 

Figures 27 and 28 were constructed to the following scales: 
Linear, i inch = 4 feet; load, i inch = 150 pounds. 

The data of the beam and loads were: Length of beam = 
12 feet = 3 inches to scale; di = 2 feet = 0.5 inch to scale; 
^ = 3 feet = 0.75 inch to scale; d^ = 6 feet = 1.5 inches to 
scale; ri = $ feet = 1.25 inches to scale; ^2 = 7 feet = 1.75 
inches to scale; Wi = 60 pounds = 0.4 inch to scale; W2 = 45 
pounds = 0.3 inch to scale; Wz =^ 90 pounds = 0.6 inch to 
scale. 

The support reactions were first found as follows: 

Taking moments about the left support, we have 

R2 {ri + n) = Wz (n + dz) + W2 {ri + ^2) + Wi (ri - di), 
or 12 i^2 = 90 X II + 45 X 8 + 60 X 3, 

whence R2 = 127.5 pounds. 

Therefore 2?i = 60 + 45 + 90 — 127.5 = 67.5 pounds. 

Reduced to scale the reactions Ri and R2 measure 

-^ = 0.4 c; inch and — — = 0.85 inch respectively. 
150 150 

The bending-moment diagram, Fig. 29, was constructed on the 

base line m^n' to a scale such that i inch in depth represents 

300 pounds-feet. The bending moments at Wi, W2, and Ws 

were calculated as follows: 

202 ^ 

Mwi= Ri (^1 - ^1) = 67.5 X 3 = 202.5 pounds-feet = 

300 

= 0.675 i^c^ ^^ scale. 
Mw,= Ri {ri + 4) - Wi {di + ffc) = 67.5 X 8 - 60 X 5 

= 240 pounds-feet. = -^ = 0.8 inch to scale. 
300 



BENDING-MOMENT AND SHEAR DIAGRAMS 



33 



= 67.5 Xii— 60X8 — 45X3 = 127.5 pounds-feet. 

JO J t^ 

= " = 0.42 ^ inch to scale. 

Ordinates equal in length to the scale measurements repre- 
senting these bending moments were erected on m^n\ as shown, 
and their extremities joined by the broken line m'fghkn\ The 
ordinate of the resulting diagram beneath any section of the beam 
is the scale measurement of the bending moment at the section. 
Thus the ordinate under the section at a measures f f of an inch, 
corresponding to a bending moment of |-|X 300 = 217.5 pounds- 
feet. 

w 



m 



Fig.S 



R2 







Fig.31 









W 



Fig. 32. 

17. Simple Beam with Concentrated Load at Middle. — The 
beam of Fig. 30 is supported at the ends and has a load W 
concentrated at its middle. 



34 THE ELEMENTS OF MECHANICS OF MATERIALS 

W 

The reaction at each support is — The bending moment 

2 

at any section between the left support and the middle and dis- 

Wx 
tant X from the support is M = ? and increases directly as 

2 

X increases. At the middle, x = —■> and M = At any sec- 

2 4 

tion between the middle and the right support and distant xi 
from the left support, 



M 



Wxi „./ L\ Wxi ... ^WL W .. . 

2 \ 2/ 2 2 2 



and decreases as x increases, becoming o when X\ = L. The 
maximum bending moment is therefore at the middle and is 

equal to Since the equations M = and M = — {L — X\) 

4 2 2 

are those of straight lines, the bending-moment diagram has the 
triangular form shown in Fig. 31. 

W 
Commencing at the left support the shear is — ■ and remains 

2 

unchanged until directly after passing the middle section, when 

W W 

it becomes W = ' changing sign at the middle section. 

2 2 

The shear diagram is then as shown in Fig. 32. 

18. Simple Beam Uniformly Loaded. — The beam of Fig. 33 

is the same as that of Fig. 30, but the load PF, instead of being 

concentrated at the middle, is uniformly distributed over the 

whole length of the beam with w pounds per unit of length, so 

that wL, the whole load, is equal to W. 

It is evident that the support reactions are each — pounds. 

2 

To find the bending moment at any section a, distant x from 
the left support, we may assume the uniform load to be made 
up of a number of parallel forces each equal to w. There are 



BENDING-MOMENT AND SHEAR DIAGRAMS 



35 



wx of these forces between section a and the left support, and 
we may substitute for them their resultant, wx, acting at their 
center of gravity, which is midway between the section and the 
support. 




Fig. 35. 



The bending moment at the section a is then 



,, o K^^ "^Lx wx^ 
Ma = Kix — wxX- = 

2 2 2 



— {L — X) = 

2 2 



That is, the bending moment at the section is proportional to 
the product of the segments into which the section divides the 
beam, and the bending-moment diagram is therefore a parabola, 
as shown in Fig. 34, having its axis vertical and under the middle 
of the beam; for it is a property of the parabola that if a diameter 
be drawn to intersect a chord the product of the segments of 
the chord is proportional to the length of that part of the diam- 



36 THE ELEMENTS OF MECHANICS OF MATERIALS 

eter included between its vertex and its point of intersection 
with the chord. 

The maximum bending moment is evidently at the middle of 

ID T OC IDOC 

the beam, so if in the general expression Ma = for 

2 2 

the bending moment we let it: = — ' we get 

2 

,-. wL? wD wD WL 

^- = T~~^~T = ^' 

in which W = wL is the whole weight. A comparison of this 
result with that obtained for the maximum-bending moment in 
Art. 17 discloses the fact that a load concentrated at the middle 
of a simple beam occasions a bending moment twice as great 
as that due to the same load uniformly distributed. 

The shear diagram of Fig. 35 may readily be constructed. 
Denoting the shear by V, we shall have for any section distant 
X from the left support, 

V = Ri — wx = — — wx, 

2 

which is the equation of a straight line, the origin being at the 

left support. It is seen that V has its maximum value, — ' 

2 

when X = o, and that it decreases as x increases until, when 

X = — ' it becomes o. For values of x greater than — the value 

2 2 

of V becomes negative, and when x = L the value of V becomes 

The straight line whose equation is F = wx is 

2 2 

therefore rs. 

19. Dangerous Section. — The maximum bending moment is 
the controlHng influence in the design of beams, and the section 
at which it occurs is known as the dangerous section. 

It has been shown, Art. 18, that in the case of a uniformly 
loaded beam the bending-moment curve is paraboHc, and that 



BENDING-MOMENT AND SHEAR DIAGRAMS 37 

therefore the ordinate representing the bending moment is a 
continuous function of x. The determination of the dangerous 
section would then be to find that value of x which would make 
the bending moment a maximum. To do this we would place 
the first it-derivative of the bending moment equal to zero, 
and the resulting value of x would be the abscissa of the dan- 
gerous section. 

For example, the general expression for the bending moment 
of the uniformly loaded beam of Fig. 33 is 

2 

Then —— = Ri — wx, 

ax 

-r, , Ri wL L 

Ri — wx = o, whence x = — = = — • 

W 2W 2 

The maximum bending moment in this case, and therefore the 
dangerous section, is at the middle of the beam. 

If, in the general expression wx for the shear in Art. 18, 

2 

we substitute for x the value — , the result is zero, from which 

2 

we infer that in the case of uniform loading the maximum bend- 
ing moment occurs at the section where the shear is zero. 

In the equation—— = Ri — wx it will be observed that the 

ax 

second member is the general expression for the shear over the 
whole beam, from which we may also infer that the first deriva- 
tive of the bending moment is the shear. 

If, however, in addition to the uniformly distributed load the 
beam were subjected to one or more concentrated loads, or if 
there were no uniform load and the beam were subjected simply 
to one or more concentrated loads, the ordinate representing the 
bending moment would no longer be a continuous function of 



38 



THE ELEMENTS OF MECHANICS OF MATERIALS 



:*;. The first derivative of the bending moment at any section 
would, however, still be the shear at that section; but the shear 
at the dangerous section might not be zero, because the shear 
is not necessarily zero at any section. If the construction of 
the shear diagram shows that the shear is not zero at any sec- 
tion, it will also show that at some one or more sections the 
shear suddenly changed sign, or passed through zero, and the 
maximum bending moment will be found to occiu: at one of 
these sections. 

20. Simple Beam with two Equal and Symmetrically Placed 
Loads. — The beam of Fig. 36 is supported at the ends and has 
two equal and symmetrically placed loads. 




At the section under the load nearest the left support, M = 
Wa ; at any section between the loads and distant x from the left 
support the bending moment is M^ = Wx — W (x — a) = Wa; 
Sit any section between the second load and the right support 



BENDING-MOMENT AND SHEAR DIAGRAMS 39 

and distant Xi from the left support the bending moment is 
M,,= Wxi -W{xi-a) - W[xi -(L- a)] = WL- Wxi. For 
Xi = L in the last equation we get M^^ = o. From these expres- 
sions for the bending moments the diagram of Fig. 37 was con- 
structed. 

Commencing at the left support the shear is W and is unchanged 
until the first load is passed, when it becomes W — W = o. 
There is no further change until the second load' is passed, when 
the shear becomes W — W — W=— W, and so continues to the 
right support. The shear diagram is, then, as shown in Fig. 38. 

21. Bending-Moment and Shear Diagrams of Cantilevers. — 
The cantilever of Fig. 39 has a concentrated load W at the end, 
a concentrated load Wi at an intermediate point between the 
end and the support, and a uniformly distributed load of w 
pounds per foot over a portion of its length. The construction 
of the bending-moment and shear diagrams of this beam will 
serve to illustrate the three cases of a cantilever: (a) Loaded at 
the end. (b) A concentrated load at some point between the 
end and the support, (c) Loaded uniformly with w pounds per 
unit of length. 

From our definition of a bending moment it will be seen that 
the bending moment is o at the free end and a maximum at the 
wall; and the tendency being to bend the beam concave down- 
ward all the bending moments are negative. 

In the construction of the bending-moment diagram of Fig. 40 
the following data were used 

Linear scale, 0.2 inch = i foot; bending-moment scale, i inch 
in depth = 800 pounds-feet. L = 10 feet = 2 inches to scale; 
fi = 7 feet = 1.4 inches to scale; ^2 = 4 feet = 0.8 inch to scale; 
IF = 32 pounds; Wi = 44 pounds; w = 4S pounds per foot run. 

If the beam were loaded only with W at the extremity the 
bending moment at the wall would be 
— WL = — 32X10 = — 320 Ibs.-ft. = — t^^ = — 0.4 inch to scale, 



40 



THE ELEMENTS OF MECHANICS OF MATERIALS 



which would be the length of the ordinate no at the wall repre- 
senting the bending moment due to W^ and mno would be the 
complete bending-moment diagram. 



|w yh 


/// 


>v 


r'^ „ r.^ .. .. 


... . /// 


^ 


Vn^v ~^- v<'V " ^'"■"" ">r 


w lbs. .per ft. /// 




u^^ 


1 1 


[ .2 '^///// 


1 :Ei:g,39 1 1 

1 1 r 


Wg ///// 

i -T^ '^yyv/j 


I r — —r 


^j. ^vZ/ 


1 i i 
i 


7 6 5 4 3 2 1 t- 




1 ! \ 


1 1 1 i ~! \yx^ " 


"" ' 


1 i H 






'^^j-^-^^i -n 






1 i""^' 


-~- o- 


^• 


1 1 ^^ 


>v. "n '■ 






"V^ lO 




1 Kg-40 1 1 


^^\^ ?o 






^^^>v^ > 


<. 




-- . fl- 




1 




N 








^^ 


i 1 1 


\; 






_.._^ 




.1... ..L i 


1 1 




t 'g 1 




© 1 


- 




t 


\ 




* 


\ 






\ 


3 


"ci 


\ |»3 






\ f 
\ 
\ 
\ 
\ 




^. 


\ 


■■^ 



Fig. 41. 
The addition of the concentrated load Wi would occasion an 
additional bending moment at the wall of 
— WiTi = — 44 X 7 = — 308 = — t^f = — 0.385 inch to scale, 



BENDING-MOMENT AND SHEAR DIAGRAMS 41 

which would be the length of the ordinate np at the wall rep- 
resenting the bending moment due to Wi. Drawing the line sp 
we would then have mspo as the complete bending-moment 
diagram due to W and Wi. 

The addition of the uniform load occasions a bending moment 
at the wall of 

— ze;r2X-= — 48X4X2 = — 384 pounds-feet 

2 

= — If t = — 0.48 inch to scale, 

which would be the length of the ordinate ot at the wall repre- 
senting the bending moment due to the uniform load, taking 
mo as a base line. The locus of the extremities of the ordinates 
representing the bending moments at all other sections of the 
uniformly loaded part would be the parabola having its apex 
at V and passing through t. 

The complete bending-moment diagram for all the loads is 
therefore msptv. This can be checked by finding at once the 
bending moment at the wall due to all the loads, thus: 

M= -WL- Win- W2 X - = - 32 X 10 - 44 X 7 - 48 X 4 X 2 

2 

= IOI2 pounds-feet = Vo¥ = 1.265 inches to scale. 

The part of the bending-moment diagram due to the uniform 
load being negative properly belongs below the base line mo, as 
shown by the dotted lines, but it has been placed above mo in 
order to obtain a solid diagram from which the bending moment 
at any section under the uniform load may be obtained by one 
measurement. 

In the construction of the shear diagram of Fig. 41 the load 
scale was taken as i inch =160 pounds. 

If W were the only load on the beam the shear throughout the 
beam would be — IF = — 32 pounds = - t¥o =0.2 inch to scale. 
Between the point of application of Wi and the wall the shear is 
increased by the amount — Wi = — 44 pounds = — tVo = — 0.275 



42 THE ELEMENTS OF MECHANICS OF MATERIALS 

inch to scale. The addition of the uniform load increases the 
shear uniformly from zero at its commencement to — wr^ pounds 
= — 48 X 4 = — 192 poimds = — lit = — 1.2 inches to scale at 
the wall. 

This can be checked by finding at once the shear at the wall 
due to all the loads, thus: 

Shear at wall = — W — Wi — wr2 = — 32 — 44 — 48 X 4 = — 268 lbs. 
= — fif = — 1-675 inches to scale. 

22. Beam with Overhanging Ends and Uniformly Loaded. — 
The beam of Fig. 42 is uniformly loaded with w pounds per unit 
of length and overhangs the supports a distance a at each end. 
Consider, quite independently of each other, the two overhangs 
as cantilevers and the part between the supports as a simple 
beam. 

Since the overhangs tend to bend concave downward the 
bending moments due to them are negative. The bending 
moment at each support due to the overhanging cantilevers is 

— wa X - = and, as we have seen in Art. 21, the para- 

2 2 

bolic curves joining the extremities of the beam with the extremi- 
ties of the ordinates representing the bending moments at the 
supports form the bending-moment diagrams of the overhangs, 
Fig. 43. Disregarding the load on the central span of the beam, 
the reactions at the supports are each wa. Then the bending 
moment for any section between the supports and distant x from 
the left support is 

M = wax — wai- -\- xj = ? 

showing the bending moment due to the overhanging loads to 
be constant between the supports and negative in character. 
Therefore the negative part below the base line mn of Fig. 43 
represents the bending moments of the whole beam due to the 
overhangs. 



BENDING-MOMENT AND SHEAR DIAGRAMS 



43 




Fig. 45- 



44 THE ELEMENTS OF MECHANICS OF MATERIALS 

The uniformly loaded central span of length h occasions a 

wh 
reaction at each support of — ? and as the tendency is to bend 

2 

the beam concave upward the bending moments are positive. 
The bending moment at any section between the supports and 
distant x from the left support is 

,, whx wx^ wx n V 
M = = — {h-x), 

2 2 2 

which, as we have seen, is the equation of a parabola having its 
axis vertical over the middle of the beam.' When x = - we have 

2 

for the maximum bending moment 

The parabola above the base line of Fig. 43 is the bending-moment 
diagram of the part of the beam between the supports due alone 
to its load, and is positive. By superposition the resiiltant 
bending-moment diagram of the whole beam. Fig. 44, is 
obtained. 

Considering the beam in its entirety, the shear of the overhang 
at the left end increases imiformly from zero at the extremity to 
—wa at the left support. The reactions due to the load are 

each wa -{- — ? so that at the instant of passing the left support 

2 

the shear becomes 

, wb wh 

— wa -\- wa -] = — • 

2 2 

At the middle the shear becomes 

, wb wb 

— wa -\-wa -\ = 0. 

2 2 

Immediately to the left of the right support the shear is 

, wb , wb 

— wa -\-wa-\ wo = — — 

2 2 



BENDING-MOMENT AND SHEAR DIAGRAMS 45 

Just passing the right support the shear is 

— wa -f-wa-i wo -{- wa -\- — = wa. 

2 2 

At the extreme right end the shear is 

. wb , , . wb 

— wa -hwa -\ wo -\- wa -\ wa = o, 

2 2 

These values of the shear enable the shear diagram of Fig. 45 to 
be constructed. 

23. Simple Beam Loaded Uniformly over a Part of its Length 
Adjoining one Support. — The beam of Fig. 46 is supported at 
the ends and loaded with w pounds per unit of length for a dis- 
tance r from the left support. 

To find the support reactions we take moments about the 
supports, thus: 

LRi = wr{L jj whence Ri = wrli =J; 

WT^ - Wf 

LR2 = — ' whence R2 = — r* 
2 2 L 

The general expression for the bending moment within the 
limits of the loaded part of the beam is 

,2 



For X = rwe get 



,, o WX' 

M = Rix 

2 






as the bending moment at the section where the load ceases. 

On the base line ab, Fig. 47, erect the ordinate cd under the 

section of the beam at the right extremity of the loaded part, 

wv^ ( t\ 
and make it equal in length to — (i — yjtoa chosen bending- 

moment scale. 



46 



THE ELEMENTS OF MECHANICS OF MATERIALS 



The general expression for the bending moment at any section 
between the loaded part and the right support is 




the equation of a straight line which gives a zero value for M 
when Xi = L. The triangle hcd, therefore, is the diagram of the 
bending moments for the unloaded part of the beam. 



BENDING-MOMENT AND SHEAR DIAGRAMS 47 

The maximum bending moment will evidently occur at some 
section within the limits of the load. To locate this section we 
proceed thus: 

Within the Hmits of the load 

2 

—— = Ki — wx, 
dx 

Ri — WX = o, whence x = —- 

w 

The maximum bending moment therefore occurs at the section 

distant — from the left support. Substituting this value of x 
w 

in the general expression for the bending moment within the 

limits of the load we get 

M -El_ -^^ - ?l 

-'■yJ- max — 9 

W 2'Ur 2W 

Under the section distant — from the left support erect the 

w 

R ^ 
ordinate ef and make it equal in length to — ~ to the chosen 

2W 

scale. This ordinate is the axis of the parabola representing 
the bending moments due to the uniform load, and since this 
parabola must pass through the points a and d it is readily con- 
structed as shown, and the bending-moment diagram of the 
beam completed. 

The general expression for the shear over the loaded part of 
the beam is Ri — wx, which, for x = o, gives Ri as the shear 
at the left support. For x = r the shear becomes Ri — wr 

(T \ WT 
I r ] — wr = = — R2 and remains unchanged 
2 L/ 2 L 

over the unloaded part of the beam. From the shear values just 

found the shear diagram of Fig. 48 is constructed, and it will be 



48 



THE ELEMENTS OF MECHANICS OF MATERIALS 



observed that the maximum bending moment occurs at the 
section where the shear is zero. 




24. Simple Beam Loaded Uniformly over a Part of its Length 
not Adjoining Either Support. — The beam of Fig. 49 is sup- 
ported at the ends and loaded with w pounds per foot for a 



BENDING-MOMENT AND SHEAR DIAGRAMS 49 

distance r commencing at a distance a from the left support and 
terminating at a distance h from the right support. 

To find the support reactions we take moments about the 
supports, thus: 

LR\ = wr [h -\- -\ whence i?i = — (& + -]; 

LRi = wrfa -i- -)> whence R2 = -ria -\- -)• 

The general expression for the bending moment between the 
left support and the section where the loading begins is 



M = R,x.=^^(b + ^^, 



the equation of a straight line. For Xi = o, M = o; ior Xi = a, 
M = —f-y^ + ")• 0^ ^^ b^s^ li^^ ^^j Fig- 50? ^^^ under the 
section distant a from the left support, erect the ordinate cd, 
and make it equal in length to — — ( ^ + -) to some chosen scale 

for bending moments. Join m and d and the triangle mcd is 
then the diagram of the bending moments of the imloaded part 
of the beam adjoining the left support. 

The general expression ior the bending moment of the un- 
loaded part of the beam adjoining the right support is, taking 
moments to the right, 

the equation of a straight line. For X2 = a -]- r, 

and ior 0C2 = o, M = o. 
At a distance oi a -\- r from the left support erect the ordinate 

gh and make it equal in length to wrla + Ali —]• It 



50 THE ELEMENTS OF MECHANICS OF MATERIALS 

will represent the bending moment at the section under the right 
extremity of the load. Join h and n and the triangle ngh is the 
diagram of the bending moments of the unloaded part of the 
beam adjoining the right support. 

The maximum bending moment evidently occurs at some 
section within the limits of the loaded part of the beam. To 
locate this section we proceed thus : 

Within the limits of the load 

ID 

M = RiXs fe — aY, 

2 

dM ^ . . 

dxz 

■D 

Ri — w (xs — a) = o, whence Xs = — + a. 

w 

The maximum bending moment occurs, therefore, at the section 

distant —^-\-a from the left support. Substituting this value 

w 

of Xs in the general expression for the bending moment within 
the limits of the load we get 



Mr 



Ri^ , „ w/Ri , V r> /^i , - \ 

= — ^ + Ria i—^ + a-a] = Ri(—^'\-aY 

W 2 \W I \2W I 



7? 

Under the section distant —-\- a from the left support erect 

w 

the ordinate ei and make it equal to RA — ^ + a) to the chosen 

\2W I 

scale. This ordinate is the axis of the parabola representing 
the bending moments due to the uniform load, and since this 
parabola must pass through d and h it is readily constructed as 
shown, and the bending-moment diagram of the whole beam 
completed. 

The shear over the unloaded part of the beam adjoining 
the left support is i?i, and over the loaded part of the beam 
it is Ri — w {x3 — a), which, for x^ = a -\- r, gives for the 



BENDING-MOMENT AND SHEAR DIAGRAMS 51 

shear at the section at the right extremity of the load, Ri — wr 

=f(' + 9-='r('+;-^)=-x(-+-:)=-''-- 

continues unchanged to the right support. From these shear 
values the shear diagram of Fig. 51 is constructed, and it will 
be observed that the maximum bending moment occurs at the 
section where the shear is zero. 

Example. — The beam of Fig. 52 is 32 feet long and overhangs 
the right support 6 feet and the left support 8 feet. In addition 
to a uniformly distributed load of 20 pounds per foot it carries 
concentrated loads of 200 pounds and 390 pounds as shown. 
Find the maximum positive and negative bending moments 
and construct the bending-moment and shear diagrams. 

Solution. — The support reactions must first be found by 
taking moments about the supports; thus, 

18 R1 + 20 X 6 X 3 = 20 X 26 X 13 + 200 X 22 + 390 X 6, 
whence Ri = 730 pounds. 

18 i?2 + 20 X 8 X 4 + 200 X 4 = 20 X 24 X 12 + 390 X 12, i 
whence R2 = 500 pounds 

In complicated problems Hke this it is the best practice to 
calculate the bending moments at various sections of the beam 
and use the results, when reduced to scale, as the lengths of 
ordinates in the construction of the bending-moment diagram. 
In this instance the linear scale will be taken as o.i inch = i foot, 
and the bending-moment scale as i inch = 1000 pounds-feet. 

For the purpose of constructing the bending-moment diagram 
these bending moments are obtained by calculation 

At the section under Wi 

M=— 20X4X2 =—160 pounds-feet. 

At the left support 

M=— 20X8X4 — 200 X 4 = — 1440 pounds-feet. 



52 THE ELEMENTS OF MECHANICS OF MATERIALS 




BENDING-MOMENT AND SHEAR DIAGRAMS 53 

At the section 2 feet to the right of the left support 

M = 730 X2 — 20X10X5 — 200 X 6 = — 740 Ibs.-ft. 

At the section 4 feet to the right of the left support 

M = 730 X4 — 20X12X6 — 200 X 8 = — i2olbs.-ft. 

At the section 6 feet to the right of the left support 

M = 730 X6 — 20X14X7 — 200 X 10 = 42olbs.-ft.' 

At the section 10 feet to the right of the left support 

M = 730 X 10 — 20 X 18 X 9 — 200 X 14 = i26olbs.-ft. 

At the section under W2 

M = 730 X 12 — 20 X 20 X 10 — 200 X 16 = 1560 Ib.-ft. 

At the section 4 feet to the left of the right support 

M = 730 X 14 — 20 X 22 X II — 200 X 18 — 390 X 2 
= 1000 pounds-feet. 

At the right support 

-^ = 730 X 18 — 20 X 26 X 13 — 200 X 22 — 390 X 6 
= — 360 pounds-feet. 

At the right end 

M = 730 X 24 — 20 X 32 X 16 — 200 X 28 — 390 X 12 
-f 500 X 6 = o. 

Reducing these bending-moment results to scale by dividing 
each by 1000, and using the results as the lengths of ordinates 
with respect to mn as an axis, the bending-moment diagram of 
Fig. 53 was constructed. 

The maximum positive bending moment is seen to be at the 
section under W2, the ordinate measuring 1.56 inches, which, 
reduced to scale, represents 1560 pounds-feet. The maximum 
negative bending moment is under the left support and measures 
1440 pounds-feet. The bending moment which is numerically 
greatest, whether positive or negative, is the one to be considered 
in the design of the beam. 



54 THE ELEMENTS OF MECHANICS OF MATERIALS 

The shear diagram of Fig. 54 may be obtained by taking the 
algebraic sum of the forces to the left of different sections, thus : 

Commencing at the left end, the shear increases uniformly and 
negatively at the rate of 20 pounds per foot from the value zero 
until, at the section immediately to the left of Wi, it becomes 

— 20 X 4 = — 80 pounds, 

which, to a scale of 800 poimds to the inch, is represented in 
the diagram of Fig. 54 by — o.i inch. 
Just passing Wi the shear becomes 

— 80 — 200 = — 280 pounds, 

which, to scale, is represented by — 0.35 inch. 

At the section just to the left of the left support the shear 
becomes 

— 20 X 8 — 200 = — 360 pounds, 

which, to scale, is represented by — 0.45 inch. 
Just passing the left support the shear becomes 

— 20 X 8 — 200 + 730 = 370 pounds, 

which, to scale, is represented by 0.4625 inch; and so on to the 
right end of the beam. 

The shears at the different sections may be obtained also by 
taking the first derivatives of the bending moments at the 
sections, thus: 

Reckoning from the left end of the beam, the expression for 



the bending moment for any section up to Wi is — 



20 x^ 



2 



Then -r~ =— 20:^=— 20X4= — 80 pounds 

ax 

for the shear just to the left oi Wi. 

For any section between Wi and the left support 

M = 200 {x — 4). 



BENDING-MOMENT AND SHEAR DIAGRAMS 55 

Then — - = — 20 a: — 200 = — 20 X 4 — 200 = — 280 pounds 

ax 

for the shear at the section just to the right of Wi\ and 

— — = — 20 a: — 200 = — 20 X 8 — 200 = — 360 pounds 
ax 

for the shear at the section just to the left of the left support. 
For any section between the left support and W2 

M = Ri{x-%) -Wiix-4)- ^^ 

2 

Then — - = Ri — Wi — 2ox = 730 — 200 — 20 X 8 = 370 lbs. 

ax 

for the shear at the section just to the right of the left support; 
and so on to the right end of the beam. 

The completion of the shear diagram by either method is left 
as a study for the student. 

25. Moving Loads. — A load moving over a structure, such 
as a train moving over a bridge, is a live load, sometimes known 
as a rolling load. A study of the bending moment and shear 
efifects of the different conditions of loads moving over structures 
is beyond the scope of this work, and only the simple case of a 
single concentrated load moving over a beam will here be given. 
For an extended study of 'moving loads recourse must be had 
to a more advanced treatise on the Mechanics of Materials. 

26- Simple Beam with a Concentrated Moving Load. — Ne- 
glecting the weight of the beam of Fig. 55, suppose the load W 
to move from left to right. 

W (L — x) 
By moments about the right support we find Ri = — ^^ • 

While the load is between the left support and some section a 
distant r from the support, the bending moment is 

M = Rir-W(r-x) = ^'' ^^~ ^^ - W{r-x) = Wx{i-j)^ 

This result must be positive, since r is less than L, and will 



56 



THE ELEMENTS OF MECHANICS OF MATERIALS 



increase in value as x increases, until the load reaches a, where r 
becomes x^ and we shall have for the bending moment at the load, 



M^Wx 



(3) 




When the load passes section a and moves toward the right 
support, the bending moment at a becomes R^r and decreases 
in value, since i?i becomes less as W approaches the right sup- 
port. When W reaches the right support, Rx becomes zero 
and the bending moment at a becomes zero. The bending 



BENDING-MOMENT AND SHEAR DIAGRAMS 57 

moment at a is greatest when the load is at a, and therefore the 
bending moment at any section is a maximum when the load is 
at the section. 

To find the dangerous section, the value of x in equation (3) 
which makes M sl maximum must be found. Thus, 

dM rrr 2 Wx , L 

-—- = W — = o; whence, x = —- 

dx L 2 

The maximum bending moment at the load, and therefore at 
the dangerous section, is then at the middle of the beam. 

Since the maximiun bending moment at any section occurs 
when the load is at the section, the diagram of maximum bend- 
ing moments, Fig. 56, is constructed from equation (3), which is 

WL 

that of a parabola, the maximum ordinate being at the 

4 

middle, where x =—- 
2 

The shear diagram is shown in Fig. 57. At the instant the 

load starts to move to the right from the left support, Ri = W 

and R2 = o. As the movement continues the shear V to the 

left of the load is 

a straight-line equation. At fhe left support x = o and V = W; 
at the right support x = L and V = o. The line of equation 
(4) is then bd, and the diagram for the shear to the left of the 
load takes the triangular form bed. 

To the right of the load during the movement the shear is 

V = R,-W = -R2=-^, (5) 

a straight-line equation. At the left support x = o and F = o; 
at the right support x = L and V = — W. The line of equa- 
tion (5) is then ce, and the diagram for the shear to the right of 
the load takes the triangular form ced. 



58 



THE ELEMENTS OF MECHANICS OF MATERIALS 



27. B ending-moment Influence Line. — An ordinate of the 
bending-moment diagram of Fig. 56 represents the greatest bend- 
ing moment produced by the rolling load at the section corre- 
sponding to the ordinate, but it is obvious that as the load 
moves over the beam the bending moment at the section changes 
continuously. The graphic representation of the variation of 
the bending moment due to a rolling load at a given section of a 
beam or structure is called the influence line of the bending 
moment at the section. 




Let X denote the distance from the left support of a con- 
centrated load W moving over the beam of Fig. 58, and let 



BENDING-MOMENT AND SHEAR DIAGRAMS 59 

/ denote the distance of a given section a from the left 
support. 
While W is at the right of a we shall have 

Ma = Rir = ^— ^• 

If W be taken as the unit load of one pound or of one ton we 
shall have 

^^^rjL-x)^ (6) 

the equation of the straight line mn, Fig. 59, the locus of all 
values of Ma between the limits oi x = L and x = r. 
When the load moves to the left of a we have 

Ma = Rir — W{r — x) = ^y —W{r — x) = ^ -^ 

For the unit load we shall then have 

M.^^-^. (7) 

Equation (7) is that of the straight line on^ which is the locus 
of the values of Ma between the limits x = r and x = o. 

For X = r \vl equations (6) and (7) we have the same value, 

Y (J_^ — r\ 
Ma = -^ — - — ' hence the »two lines intersect at w, and mno is 

the influence line of the bending moment at section a. The 
ordinate to the influence line under any section measures the 
bending moment at a due to the unit load when moving over 
the section, and the whole moment is found by multiplying the 
ordinate by W . Thus, the bending moment at section a when 
the load passes over section h is Wy. 

When the load comes on the beam from the right, the bending 
moment at a increases uniformly from zero to the maximum at 
n^ and then decreases uniformly to the value zero at the left 
support. 



6o THE ELEMENTS OF MECHANICS OF MATERIALS 

28. Shear Influence Line. — The influence Hne for the shear 
at section a of Fig. 58 may be expressed graphically. 

While the moving load W is to the right of a the shear V 
at a is 

7, = i?i = ^^L-^) . and for the unit loadF„ = ^^^ • (8) 

Equation (8) is that of the straight line m'n^, Fig. 60, the locus 
of the shear between the Hmits x = L and x = r. 
When the load moves to the left of a the shear is 

V^ = R^-W= -i^2= -^; and for the unit load,F«= -f • (9) 

Equation (9) is that of the straight line which is the locus of 
Va between the limits x = r and x = o. 

The slopes of equations (8) and (9) being the same, the lines 
are parallel, and o'n" parallel to n'fnf is the straight line of 
equation (9), and the broken line m'n^n"o' is the influence line 
of the shear at the section a while the load moves over the beam. 

When the unit load comes on the beam from the right the 
shear at a increases uniformly from zero to the value sn' at the 
section immediately to the right of a. As the load passes a 
the shear suddenly drops by the amount of the unit load and 
becomes negative, and then imiformly decreases nimaerically 
until it becomes zero at the left support. 

The ordinate y^ measures the shear at section a when the unit 

load passes over section &, and the shear at a due to the whole 

load is Wy\ 

PROBLEMS 

1. A uniform beam 25 feet in length, whose weight is disregarded, is 
supported at the ends and has a concentrated load of 400 pounds at 9 feet 
from the left support, and one of 500 pounds at 18 feet from the left sup- 
port. Construct the bending-moment and shear diagrams. 

2. Construct the bending-moment and shear diagrams of the beam of 
problem i, taking into consideration the weight of the beam, which is 
500 poimds. 



SHEAR DIAGRAMS 6l 

3. A beam 12 feet long is supported at the ends and loaded with a 
weight of 3 tons at a point 2 feet from one end. Find the bending moment 
and shear at the middle of the beam. Ans. 3 tons-ft.; 0.5 ton. 

4. A cantilever projects 10 feet from a wall and carries a uniform load 
of 60 pounds per foot; it also supports three concentrated loads of 100 
300, and 500 pounds at distances from the wall of 2, 5, and 9 feet respec- 
tively. Construct the bending-moment and shear diagrams. 

5. A beam overhangs both supports equally, carries a uniform load of 
80 pounds per foot, and has a load of 1000 pounds in the middle. The 
length of the beam is 15 feet, and the distance between the supports 8 
feet. Construct the bending-moment and shear diagrams. 

6. A beam with equal overhanging ends is loaded with three equal con- 
centrated loads, one at each end and one at the middle. Find the distance 
between the supports, in terms of the total length L, when the bending 
moment at the middle is equal to the bending moment over the supports. 
At what sections is the bending moment zero ? 

Ans. — ; - from each support. 

5 5 

7. A beam 32 feet long and supported at the ends is loaded uniformly 
for a distance of 20 feet from the left support with 0.8 ton per foot. Lo- 
cate the dangerous section, and find by calculation the maximum bending 
moment and the bending moments at the middle of the load, at the end of 
the load, and at the section 5 feet from the right support. Construct the 
bending-moment and shear diagrams. 

Ans. 13.75 ft. from left support; 75.625 tons-ft.; 70 tons-ft.* 60 tons-ft.; 
25 tons-ft. 

8. A beam 30 feet long and ^supported at the ends is loaded uniformly 
with 0.4 ton per foot for a distance of 10 feet, the load commencing at a 
distance of 8 feet from the left support and terminating at the section 12 
feet from the right support. Locate the dangerous section, and find by 
calculation the maximum bending moment and the bending moments at 
the extremities of the load. Construct the bending-moment and shear 
diagrams. 

Ans. 13.667 ft. from the left support; 24.56 tons-ft.; 20.8 tons-ft.; 18.14 
tons-ft. 



CHAPTER III 
THE THEORY OF BEAMS. BEAM DESIGN 

29. Stress and Strain. — In the discussion of the theory of 
beams it is important to distinguish between stress and strain; 
that is, it must be remembered that the change of form which 
a load produces in a body is called the strain, or deformation, 
of the body due to the load; and that the internal, or molecular, 
resistance which the material of a body interposes to resist 
deformation is called the stress. 

30. Coefficient of Elasticity and Hooke's Law. — Within the 
elastic limit of all materials the stresses are proportional to the 
strains, but since the same intensity of stress does not produce 
the same strain in different materials we must have some definite 
means of expressing the amount of strain produced in a body by 
a given stress. The means employed is to assume the body to 
be perfectly elastic and then state the intensity of stress neces- 
sary to strain the body by an amount equal to its own length. 
This stress is known as the Modulus of Elasticity, or the Coeffi- 
cient of Elasticity, and is denoted by E. The values of E for 
different materials have been detemiined and are practically 
the same for tension and compression. The mean values for E 
in pounds per square inch for the materials most commonly 
used in engineering are as follows: Timber, 1,500,000; cast 
iron, 15,000,000; wrought iron, 25,000,000; steel, 30,000,000. 

There are no materials of construction that are perfectly 
elastic, and but few that will stretch o.ooi of their lengths and 
remain elastic ; but within the elastic limit all bodies are assumed 
to be perfectly elastic, and if that limit be not exceeded the 

62 



THE THEORY OF BEAMS — BEAM DESIGN 63 

coefficient of elasticity is the ratio of the unit stress to the unit 
strain. This is in accordance with the discovery made in 1678 
by Robert Hooke and is known as Hookers law. This law 
expresses the fact that the ratio of the miit stress to the unit 
deformation is a constant quantity, known as the Modulus of 
Elasticity. Expressed as a formula, we have 

Unit stress ^ 

= hi. 



Unit deformation 



Cast iron, cement, and concrete are among the few materials 
that do not conform to Hooke's law, their coefficients of elas- 
ticity varying under different stresses; but it is customary in all 
investigations to assume the truth of the law, modifying results 
by factors of safety. 

In the case, of shear stress the modulus is known as that of 
transverse elasticity or coefficient oj rigidity^ commonly denoted 
by G. The mean values for G in pounds per square inch for the 
common engineering materials are: Timber, 150,000; cast iron, 
5,500,000; wrought iron, 10,500,000; steel, 12,500,000. 

31. The Determination of E. — If y denotes the strain pro- 
duced in a body by a stress 
5, and Y the strain produce^ 
by the stress necessary to 
stretch the body to double 
its length, then, since the 
stresses are proportional 
to the strains, we shall 
have from the stress-strain 
diagram. Fig. 61, 




Fig. 61. 



2- = —^ whence E = — = — ? 

YE y y 

since the strain, or deformation, F, is equal to L. 



64 



THE ELEMENTS OF MECHANICS OF MATERIALS 



32. Neutral Axis. — When a beam, Fig. 62, is subjected to a 
bending moment its axis is deflected from its original straight 
position into one of curvature, known as the elastic curve. The 
fibers in the horizontal layers on the convex side are in tension, 
and those on the concave side in compression. There must 
therefore be a layer, or surface, of fibers that is neither in ten- 
sion nor in compression, called the neutral surf ace, and the line 

A 

/i\ 




Fig. 62. 

in which this surface intersects the plane of a section of the 
beam is the neutral axis of the section. The neutral axis inva- 
riably passes through the center of gravity of a section whose 
material has its modulus of elasticity the same in tension as in 
compression. 

Consider the portion of the beam included between the 
sections AB and CD to be bent to the arc of a circle, and let r 
be the radius of the neutral surface, and c the distance from the 



THE THEORY OF BEAMS — BEAM DESIGN 65 

neutral surface to the outermost surface. The length of the 
neutral surface will not change in the bending, hence rd denotes 
the original length of the outermost surface expressed in radians, 
and (r -\- c) 6 expresses the length of the outermost surface after 
bending, hence 

Deformation = (r -{- c) 6 — rd = cd. 

But we have seen that 

Deformation _ 5 
Original length E 
hence 

cO _c ^S 
rd~ r~ E 



(10) 



^.1 



li 



That is, — = — = unit stress at the distance unity from the 
r c 

neutral surface, 5* being the unit 
stress, tension or compression, of 
the fibers at the outermost surface, 
and c the distance from the neutral 
surface to the outermost surface. 

33. Resisting Moment. — Sup- 
pose a beam section, Fig. 63, to 
be made up of a great number 
of layers a, a\, (h, etc., distant ^^s- 63. 

y, yi, y2, etc. from the neutral axis. The unit stresses at distances 

s s s 

y, yi, >'2,.etc., are - X ^'j - X yi, - X y2, and the total stresses 

5 5 5 

on the elemental areas are - X ay,- X aiyi, - X (hy2, etc. The 

c c c 

S S S 

moments of these stresses are - X ay^, — X a^yx", - X (hy'i, etc. 

c c c 

The sum of all these moments will be the resisting moment of 

the section, hence 

5 SI 

Resisting moment = - {ay"^ + aiy^ + a2y2^ + etc.) = — j 
c c 



66 THE ELEMENTS OF MECHANICS OF MATERIALS 

in which / is the moment of inertia of the section about the 
neutral axis. The bending moment must be equal to the resist- 
ing moment, since for equilibrium the internal stresses must 
be equal to the external forces; therefore 

SI 

Bending moment = M = (ii) 

c 

SI . 
The equation M = — is the fundamental equation for beam 
c 

investigations, and since / is expressed in bi-quadratic inches, 
M must be expressed in pounds-inches. "The length of a re- 
quired beam and the load to which it is to be subjected being 
known, we can readily find the maximum bending moment, and 
then by assuming the allowable working stress, S, per unit of 

M I 
area at the outermost fiber, we shall have — = — • The numeri- 

i3 C 

cal value thus obtained for - must be satisfied by the dimensions 

c 

of the cross section of the beam. The smaller value of S, whether 

for tension or compression, should be taken. 

34. Section Modulus. — The factor - contains the dimen- 

c 

sions of the section and is a comparative measure of its strength. 

It is called the modulus of the section and is denoted by Z. 

From equation (11) we have 

M I ^ . ^ M 

— = - = Z, whence o = — • 

Substituting this value of S in equation (10), we have 

c M . ,, EZc EI , . 

- = — - > whence M = = — y (12) 

r EZ r r 

which is the equation of the elastic curve, showing the relation- 
ship between the bending moment at any section and the radius 
of curvature of the beam. 



THE THEORY OF BEAMS — BEAM DESIGN 67 

35. Assumptions in the Theory of Beams. — In the theory of 
beams there are three important assumptions made: (a) That 
the strain increases directly as the distance from the neutral 
axis; {b) that the stress varies as the strain; (c) that the coeffi- 
cient of elasticity is the same for tension as for compression. 

No one of these assumptions is perfectly true, but for nearly 
all materials they are so nearly true that for practical purposes 
they may be regarded as true, so long as the elastic limit is not 
exceeded. 

36. Beam Design. — Beams are divided into two general 
classes: Those of imiform strength and those of uniform cross 
section. 

Beams of uniform strength are so shaped that the unit stress 
in the surface fiber is the same at all cross sections. They are 
employed when it is desired to reduce to a minimum the weight 
of the material in a beam. In their design the section modulus 
must vary directly with the bending moment, since the unit 
fiber stress remains constant. 

Beams of uniform cross section are those ordinarily used, as 
they are readily sawed from wood or rolled from steel. In such 
beams the section modulus remains constant and the fiber stress 
varies directly with the b ending moment. 

A beam is designated by its depth, and its design is a 
question of determining the sectional dimensions which will 
bear mth safety a given load, the length of the beam being 
known. 

If the dimensions of the beam are to be determined from the 

maxim-um allowable fiber stress, the bending moment at the 

dangerous section must be computed, after which the section 

I . SI . . 

modulus, - , may be found from the equation M = If it is 

c c 

desired to use one of the standard types manufactured by the 

different steel companies, the beam in the manufacturers' tables 



68 THE ELEMENTS OF MECHANICS OF MATERIALS 

having an - equal to or next greater than the one found is to be 
c 

selected. 

The standard I beams so extensively used in structural work 
are rolled in light, intermediate, and heavy weights of thirteen 
sizes. The different manufacturers issue handbooks containing 
tables of the properties of the beams they produce, and while 
there is an agreement in sizes as to depth, there are marked 
differences in the proportions of cross sections, and therefore 
also of weight per foot and of moments of inertia. 

The upper and lower horizontal parts of an I section are called 
the flanges J and the vertical part connecting them is called the 
weh. In the design of an I beam the web is assumed to 
resist the vertical shear, the flanges alone resisting the bending 
moment. 

For the support of loads beyond the capacity of a single I beam, 
two or more of them may be bolted together side by side, in 
which case cast-iron separators are used, fitting neatly between 
the flanges and bolted to the webs so as to unite the beams 
forming the girder and cause them to act together in resisting 
the load. These separators should be placed at each end of the 
girder and at points where loads are concentrated. For uniform 
loading the spacing of the separators should not be greater than 
twenty times the width of the smallest beam flange in order 
to prevent failure by buckling at the upper flanges, which are 
in compression. 

For the support of still greater loads a useful and economical 
section can be formed by the combination of two I beams 
having plates riveted to their top and bottom flanges, forming 
what is known as a hox girder. 

Should the cross section of a beam be other than those listed 
as standard, its dimensions may be determined by first computing 
the maximum bending moment and then, knowing the safe 



THE THEORY OF BEAMS 



BEAM DESIGN 



69 



allowable value of S, one dimension is assumed and the other 
computed. 

Example I. — A yellow pine beam of 18 feet span is to support 
concentrated loads of 1200 pounds and 1500 pounds at points 
4 feet and 10 feet from the left support. Taking the ultimate 
tensile strength of yellow pine as 9000 pounds per square inch, 
determine the cross-section dimensions of the beam. 



1200 lbs. 






-18- 



-10^ 



m 



Ra 



Fig. 04.' 

Solution. — From Fig. 64 we have 

18 Ri = 1200 X 14 + 1500 X 8, whence Ri = 1600 pounds; 
also 

18 R2 = 1500 X 10 + 1200 X 4, whence R2 = 1 100 pounds. 

The construction of the shear diagram shows the shear to pass 
through zero at the section under the 1500-pound load, thus 
locating the dangerous section. Then 

Mmax = 1600 X 10 — 1200 X 6 = 88oolbs.-ft. = 105,600 Ibs.-ins. 
Using a factor of safety of 10, we shall have 



5 = 



9000 
10 



900 pounds per square inch 



as the fiber stress at the outermost surface. Assuming a breadth 

of 5 inches, and remembering that c = - ' we have 

2 

M 105,600 T bd^ Kd^ , J , 

— = — — = - = = ^i— , whence d = 12 nearly, 

o 900 c 12 c 6 

so that a 5-inch by 12-inch beam would appear safe. 

We shall, however, investigate the influence of the weight of 
the beam itself. 



70 THE ELEMENTS OF MECHANICS OF MATERIALS 

Yellow pine weighs 40 pounds per cubic foot, hence 

117 • 1,4- r u s X 12 X 18 X 12 X40 

Weight of beam = ^ ^ ^^^*^ =300 pounds 

1728 ^ 

= 16.66 pounds per foot. 
Then Ri becomes 1600 + 150 = 1750 pounds, and we shall 
have 

^max = 1750 X 10 - 1200 X 6 - 16.66 X 10 X 5 = 9466.66 Ibs.-ft. 
= 113,600 pounds-inches. 

rpi e ^^ 113,600 X 6 X 12 „ - 

Then S = —r- = — = 947 lbs. per square mch, 

/ 5 X 1728 y-ri r^ ^ 

which is greater than the assumed safe stress of 900 pounds, so 
a 5.5-inch by 12-inch beam will be tried. Then 

117 • -U.- f U 5.5 X 12 X 18 X 12 X 40 J 

Weight 01 beam = ^^-^ =^ = 330 pounds 

1728 

= 18.33 pounds per foot, and Ri = 1765 pounds. Then 

Mniax= 1765 X 10 - 1200 X 6 - 18.33 X 10 X 5= 9533.5 Ibs.-ft 
= 114,402 pounds-inches. Then 

^ Mc 114,402 X 6 X 12 „ , , . - 

S = —r = — = 867 pounds per square inch, 

/ 5.5 X 1728 ^ ^ ^ ^ 

a result which shows the 5.5-inch by 12 -inch beam to be safe. 

Example II. — What Cambria I beam should be selected for 
a floor bearing a load of 180 pounds per square foot, the beams 
to have a span of 25 feet, to be spaced 8 feet apart, and to have 
a maximum unit stress of 16,000 pounds per square inch? 

Solution. — 

Load on each beam = 25 X 8 X 180 = 36,000 pounds = — 

25 
= 1440 pounds per foot run. Ri = 18,000 pounds, and 

^max = 18,000 X 12.5 X 12 — 1440 X 12.5 X 6.25 X 12 
= 1,350,000 pounds-inches. 

Then - = -^ = -^ = 84.37- 

c S 16,000 



THE THEORY OF BEAMS — BEAM DESIGN 71 

A reference to the Cambria handbook shows that the 15-inch 
special I beam of 65 poxmds per foot of the Cambria Steel Com- 
pany should be selected, its tabulated modulus being 84.8. 

Example III. — What is the proper size of I beam to carry a 
load of 35,000 pounds concentrated at the middle of a span of 
25 feet, the fiber stress not to exceed 16,000 pounds per square 
inch? 

Solution. — 

-^max = 17,' 500 X 12.5 X 12 = 2,625,000 pounds-inches. 

/ M 2,62c:,ooo 

- = — = -^-^ = 164. 

c o 16,000 

This value of the section modulus would indicate that a 24-inch 
standard Cambria I beam of 80 pounds per foot should be se- 
lected, its tabulated section modulus being 173.9. Its tabulated 
moment of inertia is 2087.2. 

We shall investigate the effect of the weight of the beam itself. 

Weight of beam = 80 X 25 = 2000 pounds; therefore Ri 
= 18,500 pounds. 

Then Mnxax = 18,500 X 12.5 X 12 - 80 X 12.5 X 6.25 X 12 
= 2,700,000 pounds-inches. 

„ Mc 2,700,000 X 12 , 

- -^ = ^ = 2087.2 = '5,Soo pounds, 

which is 500 pounds less per square inch than the safe allowable 
stress at the outermost fiber. The 24-inch standard Cambria I 
beam weighing 80 pounds per foot, is, therefore, the proper 
selection. (The safe loads for beams given in the handbooks 
issued by the different steel companies include the weight of the 
beam.) 

Example IV. — Heavy 1 2-inch Cambria I beams of 20 feet 
span are to be used to support a floor bearing a uniform load, 
including the weight of the beam, of 200 pounds per square foot. 
The outermost fiber stress is not to exceed 16,000 pounds per 
square inch. Find the proper spacing of the beams. 



72 THE ELEMENTS OF MECHANICS OF MATERIALS 

Solution. — From the Cambria handbook we find for the beam 
selected, 

Section modulus = - =41. 
c 

Let X denote the distance between two consecutive beams. 
Then 

Load on one beam = 20 ic X 200 = 4000 x pounds, and 

Load per foot = = 200 x pounds, 

20 

Ri = 2000 X pounds. ' 

-^max = 2000 a; X 10 X 1 2 — 200 :*:X 10X5X12 = 1 20,000 X Ibs.-ins. 

Then - = -77' or 41 = —~ , whence x = 5.47 feet. 

c o 16,000 

PROBLEMS 

1. A rectangular beam 9 inches deep, 3 inches wide, supports a load 
of 0.5 ton concentrated at the middle of an 8-foot span. Find the maxi- 
mum fiber stress. Ans. 0.3 ton per sq. in. 

2. A square timber beam of 12-inch side weighs 50 pounds per cubic 
foot, is 20 feet long, and supports a load of 2 long tons at the middle of its 
span. Calculate the fiber stress at the middle section. 

Ans. 1037 lbs. per sq. in. 

3. Light 12-inch Cambria I beams of 20-feet span are to support a 
floor bearing a uniform load of 175 pounds per square foot. The outer- 
most fiber stress is not to exceed 16,000 pounds per square inch. At what 
distance apart should the beams be placed? Ans. 5.5 ft. 

4. Wliat safe uniform load may be placed on a wooden cantilever 5 
feet long, 3 inches wide, and 5 inches deep in order that the fiber stress 
shall be 900 pounds per square inch ? Ans. 71 lbs. per ft. 

5. What safe uniformly distributed load may be placed on a standard 
Cambria channel beam weighing 20 pounds per foot, the span being 16 
feet, the web placed vertical, and the maximum fiber stress not to exceed 
12,500 pounds per square inch? Ans. 11,150 lbs. 

6. Find the factor of safety of a heavy 12-inch Cambria I beam of 15 
feet clear span when sustaining a uniformly distributed load of 30 tons. 

Ans. 2, 

7. Compare the strength of a steel beam of rectangular section, 5 inches 
by 2 inches, with that of a 9-inch I beam of the same material, length, and 
area of section, the moment of the inertia of the I beam being 11 1.8. 

Ans. I beam is three times as strong. 



CHAPTER IV 
THE DEFLECTION OF BEAMS 

37. Deflection of Beams. — In Chapter III the question of 
beam design was considered from the standpoint of the maximum 
allowable fiber stress, but the safe load for a beam to carry may 
equally well be determined from its maximum allowable deflec- 
tion. For the protection of plastered ceilings in buildings, the 
floor beams are limited in deflection to about si-^ of their length, 
and their design, therefore, is based on considerations of deflection 
rather than of strength. 

The radius of curvature of any plane curve is given in the 
differential calculus as 



In the case of a beam the curvature is assumed to be small, and 
therefore the slope of the "tangent at any point is small. In 

other words, -7^ being very small, (-7^) is infinitesimal and may 

be neglected, and therefore 

_ I 

Substituting this value of r in equation (12) of Art. 34, p. 66, 
we have 

73 



74 



THE ELEMENTS OF MECHANICS OF MATERIALS 



which is the equation of the elastic curve for the investiga- 
tion of the deflection of beams. In any particular case the 
value of M must be expressed in terms of x, and then, after 
two integrations, the deflection, y, will be known for any value 
of X. 

38. Points of Inflection. — Simple beams that are free at 
their end supports — that is, having no restraining influence 
to keep their supported ends horizontal — bend concave up- 
ward, but beams that are fixed at their end supports — that 
is, having their ends built in so that the built-in part is 
constrained to remain horizontal — are subjected to a com- 
bined curvature of bending which is concave downward near 
the supports and concave upward toward the middle. A beam 
that overhangs its supports, such as that of the Example, p. 51, 



-^=200 



di 



-18- 



4-2- 



-6'- 



Fig. 65. 



Chap. II, reproduced in Fig. 65, is also subjected to a combined 
curvature of bending between the supports. The overhangs 
exert a constraining influence to keep horizontal the parts of 
the beam resting on the supports and thus compel a bending 
of the beam concave downward near to and inside each support. 
This is clearly indicated by the dotted curves of Fig. 65. In 
these cases of double curvature in bending, the points at which 
the contrary curves separate and at which there is no bending 
whatever, are known as points of inflection or points of contrary 
flexure. They can be found by placing the general expression 



THE DEFLECTION OF BEAMS 75 

for the bending moment within their limits equal to zero. Thus, 
in Fig. 65, it is evident from inspection that there will be a point 
of inflection between the left support and W2, and one between 
the right support and W2. The expression for the bending 
moment at any section within the limits of the left support and 
W2 is 

M = RiX-Wi(x-i-4) --{x + 8)2. 

2 

Placing this expression equal to zero, and substituting 730 for 
Ri, 200 for Wi, and 20 for w (see p. 51), the resulting quadratic 
gives a value of 4.42 for x, which shows there is a point of in- 
flection at 4.42 feet to the right of the left support. The ex- 
pression for the bending moment at any section within the 
limits of TF2 and the right support is 

M = R,xi - Wi (xi + 4) - TF2 (xi - 1 2) - - (xi + 8)2. 

2 

Placing this expression equal to zero and substituting 390 for 
W2 and 500 for R2, the resulting quadratic gives for Xi the value 
of 17.027, showing the other point of inflection to be at the sec- 
tion distant about i foot to the left of the right support. These 
points of inflection are indicated at / and u in the bending-moment 
diagram of Fig. 53, p. 52, a't which points there are no bending 
moments. 

The determination of the points of inflection is very important 
in a mechanical sense as furnishing the logical location for 
expansion joints in bridge construction. 

39. Examples in Beam Deflection. — The expressions for the 
maximum deflections of beams under several different condi- 
tions of support and loading will here be derived; those for other 
conditions may be found by similar processes. 

Example 1. Simple beam with a load W concentrated at the 
middle, Fig. 66. 



76 



THE ELEMENTS OF MECHANICS OF MATERIALS 



For any point between the left support and the middle of the 
beam, we have 

Wx 



Then 
Integrating once, 



M = 

2 

dx^ 2 
Wx^ 



EI 



dy 
dx 



+ c. 




Fig. 66. 

To find C we must know the slope at some point. At the 
middle of the beam the tangent to the elastic curve is hori- 
zontal, consequently there is no slope at that point, hence 

^ = o when x = — , therefore C = -r- • 

dx 2 10 



Then 



dx 4 i6 



Integrating the second time, we get 



Ely 



Wx' WDx 



+ C'. 



12 l6 

At the supports there is no deflection, therefore 

y = o when x = o :. C = o. 
Wx' WDx 



Then 



EIy = 



12 



i6 



The deflection is a maximimi when x = — , therefore 

2 



EIyinax = 



WD WU 



96 



Then 



^max — 



32 

WU 
4S EI 



WD 

48 



THE DEFLECTION OF BEAMS 



77 



Example II. — Cantilever with a load W concentrated at the 
end, Fig. 67. 




yw 



Fig. 67. 



For any section of the beam we have 

M ^-W{L-x) 



Then 



ax 2 

There is no slope at the wall, therefore —•= o when ic = o, 

ax 

.-. C = o. 

Then EI^ = ^^-WLx, 

ax 2 

Ely = i^ - ^^^-^^ + C'. 

6 2 



There is no deflection at the wall, therefore 3^ = when x = o, 

.'. a = 0. 

Wx^ _ WLx^ 

6 2 



Then 



Ely 



The deflection is a maximum when x = L, therefore 



Elyuiax = 



wu _ wu 

6 2 

WD 
2>El' 



WD 



78 THE ELEMENTS OF MECHANICS OF MATERIALS 

Example III. — Simple beam uniformly loaded with w pounds 
per unit of length, Fig. 6^,. For any section of the beam we have 

wLx wo^ 

2 2 



Then 



M 



-pj^y_ _ wLx wx^ 

doC^ 2 2 

pjdy _ wLx^ wx^ ^ 

dx A 6 ' 






w lbs. per unit of length 



WL 



Fig. 68. 



The tangent to the elastic curve is horizontal at the middle of 



the beam, therefore 



dy 
dx 



o when x = 



For X 



1 we have 



~ 48 ~"T6^ 



Then 



Pj-dy _ wLx- 



dx 4 
wLx^ 



Then 



EIy = 

y = 
y = 



WX'^ 

wx^ 



2 

wU 

24 

wU 

24 ' 

wDx 



+ c'. 



12 24 24 

o when x = o, consequently C = o. 

w 



24 EI 



(2 Lx^ — x^ — Ux), 



The deflection is a maximum at the middle of the beam, where 
L 



X = 



Then 3;^ 



24EI\4 16 2/ 384 £/~ 384 £/ 



y m 



which W = wL, the whole load. 



THE DEFLECTION OF BEAMS 



79 



Example IV. — Cantilever uniformly loaded with w pounds 
per unit of length, Fig. 69. 




Fig. 69. 



For any section of the beam 

w 



M = --{L- xf, 
2 



Then 



£7?-^ = 



dx'^ 



> 



dx 



dy 
dx 



o when x = o, consequently C = o. 



Then 



Then 



EI^^ = -^Uoc-Lx^ + '^\ 
dx 2\ 3 / 

^\ 2 3 12/ 

y = o when x = o, consequently C = o. 



3; =- 



{6L^x''-4Lx' + x'). 



24. EI 
The deflection is a maximum when x = L, therefore 

wL^ WU 

>'max = — '^~^T ^ ~ ^^^ ' ^^ which IF = wL is the whole load. 
8 EI 8 A/ 

Example V. — Beam built-in or fixed at the ends and uni- 
formly loaded, Fig. 70. 

This is a condition not heretofore encountered. The beam 
is fixed at the ends in the sense that the parts of the beam built 



8o 



THE ELEMENTS OF MECHANICS OF MATERIALS 



in are constrained to remain horizontal while the beam is being 
bent, but it is imder stood that the beam is free endwise. The 
reaction of the wall in keeping the built-in part horizontal intro- 
duces the couple, P, P. 




lbs. per nnit of length 




Fig. 70, 

The bending moment at any section between the left support 
and the middle, and distant x from the left support, is 



Then 



= Rix + Px- P{x + z) - 


wx^ 

2 


2 


2 


==-{Lx- x^) - Pz. 




•Z-j—^^ -'■■■• 





ax 2 \ 2 3 / 



From the nature of the support there is no deflection at the wall, 

hence -~ = o when x = o, consequently C = o. To find the 
ax 

value of Pz, which is the bending moment at the support, we 

must know the value of -~ at some other section. At the 

ax 

middle of the beam the tangent to the elastic curve is horizontal, 



THE DEFLECTION OF BEAMS 8l 

therefore -^ = o when x = -- Substituting this value of x in 
ax 2 

the first derivative, we get 

w/U D\ PzL , „ wU 

= o, whence Pz = ? 

2 V 8 24/ 2 12 

the bending moment at the support. 

Substituting the value of Pz, and integrating the second time, 
we have 

2\ 6 12/ 24 

At the support 3; = o and x = o, therefore C = o; and }; is a 

maximum when x = — ' consequently 
2 

w (U U V\ wU WD 



\48 IQ2 06/ 



-'-^ 2EI\48 192 96/ 384 £/ 384 £/ 

in which W = "ze^L is the whole load. 

Substituting the value of Pz in the expression for M, and letting 
L 



we have 
2 



M 



-'^(Ll _Ll\ .'^^^ _wL\ 



showing the bending moment at the middle to be but half that 

at the support, hence 

,, wD WL 

Mjnax = = 

12 12 

For the points of inflection we place the second derivative 
equal to zero, thus: 

- (Lx — x^) = 0, or Lx — x^ = --^ whence x = ^ , — ^ • 

2 12 6 6 

40. Continuous Beams. — The beams heretofore considered 
have been either cantilevers or beams having but two supports. 
A beam resting on more than two supports is termed a continu- 



82 



THE ele:ments of mecil\nics of materials 



ous beam. The chief difficulties in the treatment of continu- 
ous beams are the determination of the support reactions and the 
bending moments at the supports. When these are known the 
bending moment, shear, and deflection at any section may be 
determined. 

When considering beams having two supports the support 
reactions were easily found by means of two equiHbrium equa- 
tions. With parallel forces there can be but two equilibrium 
equations, so that if a beam has more than two supports some 
other means must be adopted for finding the support reactions. 
The whole treatment of continuous beams is simphfied by means 
of Clapeyron's theorem of three moments, and is fully set forth 
in any complete treatise on the mechanics of engineering. Only 
the special case of a beam uniformly loaded and resting on three 
supports equally spaced and on the same level will here be 
considered. 

41. Beam Resting on Three Supports. — The beam of Fig. 71 
is uniformly loaded with w pounds per unit of length and rests 
on three supports equally spaced and on the same level. 



Rj 



lbs, per unit of length 



R3 



Fig. 71. 



In finding the support reactions we will first assume the 
middle support to be removed. In such case we have found 
(Art. 39, Ex. Ill) that the deflection at the middle due to the 

whole load wL is-^ — --• 
384 £/ 

Let TF' denote the load on the middle support. Then the 



THE DEFLECTION OF BEAMS S^ 

a caused by the reaction at the midc 
due to the concentrated load W is, by Art. 39, Ex. I, 



upward deflection caused by the reaction at the middle support 



48 EI 

The tops of the supports being on the same level, the upward 
deflection must equal the downward deflection, so that 

-— — : = -^r—=rr- ' whence W = ^— — • 
48 £/ 384 £/ 8 

That is, the middle support bears I of the whole load; and since 
the load is uniform, each of the end supports bears t^ of the load. 

Thus Ri = R3 = ^2_^, and R2 = 

16 8 

The bending moment at any section between the left and 
middle supports, and distant x from the left support, is 

(13) 





M = 


-- Rix- 


wx'^ 
2 




Then 


-^■ 


-- Rix- 


wx^ 

2 






^'1= 


Rix' 

2 


WX^ 

6 


+ c. 


There is no slope at the middle support, 


therefore 




dy 


when 


X = 


L 

— J 
2 


consequently 











^ _ 7w^ _ Rix'^ _ wU _ 2)WD _ _ wD 
~ 6 2 ~ 48 128 ~ 384 * 

rr., T^Tdy RlX^ wx^ wU 

Then EI-^ = — ; — j 

dx 2 o 384 

6 24 384 

the constant of integration being zero, because y — o when 
X — o. 



84 THE ELEMENTS OF MECHANICS OF MATERIALS 

From equation (14) the deflection, 3;, may be determined for 
any section. Thus, at the section midway between the left and 

T "2. 11) T 

middle supports, x = — and we have, since Ri = ^—r-y 
4 16 

^ ^ wU wU' wV' wU 

2048 6144 1536 3072 

wu 

3072 £/, 

in which W = wL is the whole load. This value of y is not the 

maximum deflection, as that does not take place at the middle 

of the span. 

To find the point of inflection between the left and middle 

supports we place the expression for the bending moment within 

those Hmits [equation (13)] equal to zero, thus 

^ wLx wx^ , S L 

~ — ; = o, whence x = ^^^ • 

16 2 ' 8 

Hence there is a point of inflection distant three-eighths the 
length of the beam from the left support, or one-eighth of the 
length of the beam to the left of the middle support; there is 
another point of inflection similarly situated to the right of the 
middle support. 

To find the bending moment at the middle support, let ic = — 

2 

in equation (13) and we have 

3wL^ wV wD WL 

32 8 32 32 

The greatest bending moment between the supports will be 
at the section where the shear passes through zero. We locate 
this section by placing the first jc- derivative equal to zero; thus, 
by differentiating (13), 

-r- = Ri — WX, 

ax 

^—- WX = o, whence x = -^ ; 

16 16 



THE DEFLECTION OF BEAMS 85 

that is, the maximum bending moment between the supports 

occurs at the section distant '^-— from the left support. To find 

16 

its value we substitute ^-— for x in (13), and 
16 



M 



gwL^ _ 9^^^ _ gwL ^ _ gWL 
256 512 512 512 

This result is numerically less than that found for the bending 
moment at the middle support; hence the maximum bending 
moment is at the middle support, and we shall have for safe 

loading, — = • The bending moment and shear diagrams 

c 32 

are readily drawn, as shown in Fig. 72. 




Fig. 72. 

42. The Strength and Stiffness of Beams. — The direct meas- 
ure of the strength of a beam is the load it will safely support, 
and varies inversely as the outermost fiber stress at the dangerous 

section. Since at the dangerous section S = — , in which M 



86 THE ELEMENTS OF MECHANICS OF MATERIALS 

is the maximum bending moment, it follows that — r is the ex- 

Mc 

pression for the comparative strength of beams. The expression 

for the maximum bending moments obtained in preceding pages 

WL 

is M = ,in which w has the values i, 2, 4, 8, and 12, depending 

m 

on the conditions of support and loading. The expression for 

ifil 
the strength of a beam may then be written — — ; that is, the 

strength of a beam varies directly as the section modulus and 

inversely as the length. In the special case of beams of rec- 

hh^ h hh^ 

tangular section, where / = — and c = - , we have -— as the 

12 2 6 

value of the section modulus - , so that the strength of such beams 

c 

varies directly as the breadth and directly as the square of the 

depth. Doubling the breadth of a beam of rectangular section, 

therefore, doubles its strength; doubling the depth increases its 

strength four times. This furnishes the reason for placing 

rectangular beams with their greatest dimension vertical. 

The ratio of the maximum deflection of a beam to its length is 

termed the stiffness of the beam, and its measure is the load the 

beam can carry with a given deflection. The expression for 

WD 

nEI' 



the maximum deflections obtained in this chapter is y 



in which n has the values 3, 8, 48, '^-^, and 384, depending on 

the conditions of support and loading. From this expression 

we get W = — -~ ; hence the stiffness of a beam varies directly 

as E and /, and inversely as the cube of the length. The follow- 
ing table furnishes a comparison of beams under different con- 
ditions of loading and support, the relative strength and stiffness 
of each being expressed in comparison with that of the weakest 



THE DEFLECTION OF BEAMS 



87 



beam, which is the cantilever with a concentrated load at the 
free end: 



Kind of beam and nature of load and 
support. 


Maximum 
bending 
moment. 


Maximum 
deflection. 


Relative 

strength. 


Relative 
stiffness. 


Cantilever, load at free end 

Cantilever, load uniformly distrib- 
uted 


WL 

WL 

2 

WL 
4 

WL 

8 

WL 
12 


WL^ 
3 EI 

WU 

SEI 

WU 

48 £/ 

swu 

3HEI 

WL^ 
384 EI 


I 
2 

4 

8 
12 


I 
2| 


Simple beam, load at middle 

Simple beam, load uniformly dis- 
tributed 


16 
2Sl 


Beam with fixed ends, uniformly 
loaded 


128 







PROBLEMS 

I. Find the expression for the maximum deflection of a beam fixed at 
the ends and with a load W concentrated at the middle. Find the points 
of inflection. , WU L 



Ans. 



from each end. 



192 EI ' 4 

2. Find the maximum deflection of a 24-inch Cambria I beam, 25 feet 
long and weighing 80 pounds per foot, when resting on end supports and 
bearing a load of 35,000 pounds at the middle. Neglect the weight of the 
beam. Ans. 0.314 in. 

3. A rolled steel beam,' symmetrical about its neutral axis, has a 
moment of inertia of 72 inch units. The beam, which is 8 inches deep, is 
laid across an opening of 10 feet and carries an evenly distributed load of 
9 tons, including its own weight. Find the maximum deflection; find also 
the maximum fiber stress. 

A7ts. 0.218 inch; 7.5 tons per sq. in. 

4. A lo-inch Cambria I beam, the moment of inertia of which is tabu- 
lated as 1 2 2. 1, is laid across an opening of 16 feet. In addition to a concen- 
trated load of 1000 pounds at the middle it carries a uniformly distributed 
load of 14,400 poimds, including the weight of the beam. Find the maxi- 
mum deflection, taking E as 29,000,000; find also the maximum fiber 
stress. Ans. 0.416 in.; 16,120 lbs. per sq. in. 

5. A beam of wood, 8 inches wide and 12 inches deep, and fixed at 
the ends, covers a span of 14 feet. It bears a uniformly distributed load 



88 THE ELEMENTS OF MECHANICS OF MATERIALS 

of 10,000 pounds, including its own weight. Find the maximum deflec- 
tion, taking E as 1,500,000; find also the maximum fiber stress. 

Ans. 0.072 in.; 729 lbs. per sq. in. 

6. Find the safe uniformly distributed load for a 6-inch I beam rest- 
ing on end supports 20 feet apart if the deflection is limited to 3^7 of 
span. E = 30,000,000; I = 24; and weight of beam is 14.75 pounds per 
linear foot. Ans. 118.6 lbs. per linear foot. 

7. How much stronger is a beam 4 inches wide, 8 inches deep, and 8 
feet long, weighing 71 pounds, than one 3 inches wide, 5 inches deep, and 
14 feet long, weighing 58 pounds? Ans, 4.9 times. 



CHAPTER V 
COLUMNS. SHAFTS 

43. Columns. — A column, or strut, is a straight beam 
acted on compressively at its extremities, and of such length 
compared with its diameter, or least sectional dimension, that 
failure will result from buckling or lateral bending, instead 
of by crushing or by splitting. In addition to the many 
familiar applications of columns in structural work, the pis- 
ton rods and connecting rods of steam engines are classed as 
columns. 

If columns were initially absolutely straight, made of homo- 
geneous material, and exactly centrally loaded, there would be 
no difference in the character of their failure from that of short 
specimens. These three conditions are never fulfilled in practice, 
and in consequence columns are weaker than short blocks of the 
same material. 

No satisfactory theoretical discussion of columns has been 
made, and all the formulae used in their design contain con- 
stants determined by experiment. The formula having the most 
rational basis is the one attributed to Rankine or to Gordon, 

which is 

AS 



W 



T2 



in which W is the load on the column expressed in pounds, A the 
sectional area of the column in square inches, 5 the ultimate 
compressive strength of a short block of the material, L the 
length of the column in inches, K the least radius of gyration of 

89 



90 



THE ELEMENTS OF MECHANICS OF MATERIALS 



the section, and a a constant quantity determined by experiment 
for different materials. 

The strength of a column depends largely upon the manner 
in which its ends are secured. If the ends are flat the column 
is said to have square or flat ends ; if one end be fixed and the 
other end hinged, as in the case of a piston rod, the column is 
said to he fixed at one end d^ndfree at the other; if both ends are 
hinged, as in the case of a connecting rod, the column is said to 
be free or round at the ends. 

The value of K" in Rankine's formula is found from the rela- 
tion I = AK^. The values for S in pounds per square inch, 
for a, and for suitable factors of safety for the three conditions 
of bearing, are given in the following table: 





Timber. 


Cast iron. 


Wrought 
iron. 


Structural 
steel. 


Hard steel. 


s 


8000 


84,000 


55.000 


60,000 


120,000 


a 
(square ends) 


3000 


5,000 


36,000 


36,000 


25,000 






a 
(fixed and free) 


1690 


2,810 


20,250 


24,000 


14,060 


a 
(free ends) 


750 


1,250 


9,000 


9,000 


6,250 


Factor of safety 

(buildings) 


8 


8 


6 


4 





Factor of safety 

(bridges) 


xo 




10 


5 


7 




Factor of safety 

(shocks) 


15 




IS 


10 


I r 







It will be observed from the table that the ratios of the values 
of a for the square ends to the values for fixed and free and for 
free ends are as 1.75 to i and as 4 to i respectively, and these 
ratios indicate the relative strengths of long columns with those 
end conditions. 



COLUMNS 91 

Example I. — A hollow cylindrical cast iron column 16 feet 
long, with square ends, sustains a load of 200,000 pounds when 
used in a building. Outside diameter, 10 inches; inside diameter, 
8 inches. Is it safe? 

Solution. — 

A = 0.7854 (100 — 64) = 28.27 square inches. 

„„ I IT (10,000 — 4096) 

A 64 X 28.27 

Substituting in Rankine's formula, we have 
00,000 r (16 X 12' 

28.27 L 5000 X 10 



S = i22£22 \, + (^6Xi2)^J ^ ^^^^^g 



pounds per square inch. The factor of safety, then, is — 

= 6.9, showing the column to be unsafe. ' 

Example II. — Find the safe load 'for a 12-inch Cambria I 
column weighing 31.5 pounds per foot, 16 feet long, and mth 
square ends, the column to be used in a building. 

Solution. — Referring to the Cambria handbook, we find that 
A = 9.26 square inches, andi^ = i.oi. The factor of safety is 4. 

Then W = — - — . — r^ — = 09,300 pounds. 

(16 X 12) 

36,000 X (i.oi)^ 

44. Design of Columifs. — In designing a column we have 
given the form of its section, its length, the material of which it 
is to be made, the load it is to carry, and the manner of securing 
its ends. The problem is then to determine the necessary area 
of cross section. The general method of procedure is to deter- 
mine from the data given the necessary cross- section area for a 
short block of the material; then, knowing that the section area 
of the required column must be larger, assume dimensions which 
give a larger area, and then, by means of Rankine's formula, 
ascertain the unit stress resulting from such assumption. If it 
is too great, the assumed area is too small; if too little, a smaller 



92 THE ELEMENTS OF MECHANICS OF MATERIALS 

area must be chosen. Proceed thus by trial and error until a 
satisfactory solution is obtained. 

Example III. — A column of timber of square section, 15 feet 
long and with square ends, is required to support a steady load 
of 10 tons. Find the necessary dimensions of the section. 

Solution. — Using a factor of safety of 8, we have = 1000 

8 

poimds per square inch as the safe working unit stress. 

r^, 2000 X 10 

Then = 20 

1000 

square inches area of section needed for a very short column, or 
one less in length than ten times the least dimension of the 
section. As the required section must be larger, we will assume 
an area of 36 square inches. 

Then Z^ = l=^ = 3. 

A 12 X 36 

Therefore 

^ 20,000r , (15 X I2)n , . , 

S = — ——\ I + ^-"^ = 2555 pounds per square mch. 

36 L 3000 X 3 J 

Since this is much larger than the allowable unit stress of 
1000 pounds, we must assume a larger area. 
Try an area of 64 square inches. Then 

IC_ (8)^ _i6^ 
12 X 64 3 

^ 20,000r , (l'5 X I2)n ^ J .1 

S = —7 — I + ^-^ — — rt = 945 pounds per square mch. 
64 L 3000 XV-J 

This result is a trifle small, so we will try a square section 
whose side is 7.9 inches. Then 



^2.Jl94 = 5.,, 



12 (7.9)2 

20,000r , (15X12)21 oA J • 1- 

—2-— I + -!^^2 '— = 986 pounds per square men, 

(7.9)2 L ^3000X5-2] ^ ^ ^ ^ 



^ ^ 20,000 r^ , (15 X 12) 2 



COLUMNS 93 

which is quite near the allowable unit stress; therefore a column 
of square section, having a side of 7.9 inches, will suffice. 

Sometimes all the dimensions may be assumed except one, 
and then, after expressing A and K^ in terms of the unknown 
dimension, we can substitute them in Rankine's formula and 
solve the resulting bi-quadratic for the unknown dimension. 
Thus, in the problem just solved, let x denote the length of the 
side of the unknown square section. Then 

A = x^ and K? = r = — > 

\2%^ 12 

and 

20,000 r , (15 X 12)2 X 1 21 20,000 / , i2o.6\ 
1000 = — ^ I + ^^ —^ — = —^ — I + -~i- ' 

%^ L 3000^:1;^ J x^ \ x^ I 

whence a: = 7.9 inches. 

Example IV. — A hollow cylindrical cast iron column 20 feet 
in length is required to sustain a steady load of 164,000 pounds. 
Determine its cross-sectional dimensions. 

Solution. — Using a factor of safety of 8, we have — 

= 10,500 pounds per square inch as the safe working unit 
stress. 

Then — — = 115.62 square inches needed for a very short 

10,500 

column. 

Assuming an area of cross section of 25 square inches, and 
assuming further that the outside diameter of the column shall 
be 10 inches, we shall have, calling d the inside diameter, 

25 = 0.7854 X 100 - 0.7854 ^^ 

whence d = 8.256 inches. 

^, ^2 / x[(i o)^- (8.256) 4] 

Then K^ = -- = -^ — ^ ^ — ^-^^ = 10.51, 

A 64 X 25 ^ ' 

and 

c, i64,ooor , (20 X 12)2 "I J 

S = —^ I + -^ — '— = 13,750 pounds per sq. m., 

25 L 5000X10.51J 



94 THE ELEMENTS OF MECHANICS OF MATERIALS 

which is greater than the allowable unit stress of 10,500 pounds. 
The cross-sectional area must therefore be larger, so we will 
assume an area of 33.75 square inches. Then 

33-75 = 0-7854 X 100 - 0.7854 </^ 

whence d = 7.55 inches, 

and K' = g.^2. 

Then 

<. i64,ooor , (20X12)2"] c A • I, 

o = I + -^ 7— = 10, coo pounds per square mch, 

33.75 L 5000X9-82J '^ ^' ^ ^ 

a result sufhciently near the allowable unit stress to warrant 
making the column 10 inches in outside diameter and the metal 
1.25 inches thick. 

Example V. — Using a factor of safety of 4, what safe load 
can be supported by a 7-inch Cambria channel-and-plate column 
14 feet long and weighing 34.8 pounds per foot? 

Solution. — From the Cambria handbook the section area 

of the column is found to be 10.2 square inches, and the least 

radius of gyration 2.63. Then 

^^ ^0,000 

10.2 x^^-^ 

^ ^ AS ^ 4 10.2 X 12,500 

aK^ 36,000 X (2.63)^ 36,000 X 6.9 



= 114,500 pounds. 



PROBLEMS 



1. A hollow cylindrical cast iron column with square ends, whose 
thickness of metal is 2 inches, length 24 feet, and outside diameter 24 
inches, is to be used in a building. What safe load can it sustain ? 

Ans. 1,141,000 lbs. 

2. A round, solid, cast iron coliunn with square ends is 15 feet long 
and 6 inches in diameter. What safe load can it carry? 

Ans. 76,500 lbs. 

3. A 12-inch Cambria channel-and-plate column used in a building is 
18 feet long and weighs 64.8 pounds per foot. What safe load can it 
carry? Ans. 223,300 lbs. 



SHAFTS 95 

4. Find the external diameter and thickness of metal of a hollow cast 
iron column that will safely carry a load of 194,000 pounds, its length 
being 16 feet, and the ratio of the inside and outside diameters being 0.8. 
Use a factor of safety of 8. 

Ans. 10.5 ins.; i.i ins. 

5. Find the thickness of metal of a hollow column of structural steel, 
6 inches in diameter and 2S feet long, to support a load of 100,000 pounds 
in a building. Ans. 0.775 in. 

45. Shafts. — A shaft in mechanics is a revolving bar used 
for the transmission of power generated by an engine or other 
motor. In the design of a shaft the important consideration is 
the nature of the stress to which it is to be subjected. In the 
form of shaft commonly known as an axle, where the load is 
applied transversely, the stress is chiefly due to bending; in 
transmission shafting, such as that used in machine shops and 
factories, the stress is principally due to torsion; in heavier forms, 
such as crank shafts, the stress is one of combined torsion and 
bending. 

Since rotating cyKndrical bodies have equal strength in all 
positions, shafts are usually circular in section. We shall be 
concerned only with those shafts of circular section that are 
subjected chiefly to a torsional stress, and with those subjected 
to a combined stress o^ torsion and bending. 

46. Torsion. — When shafts are used for the transmission of 
power they are subjected to a shearing stress due to a torque or 
twisting moment, the measure of which is the product of the 
applied force and the distance of the point of its application 
from the axis of the shaft. If P denotes the applied force 
and r the perpendicular distance of its point of application 
from the axis of the shaft, then Pr is the measure of the twist- 
ing moment. 

When a uniform circular shaft is subjected to a twisting 
moment each element parallel to the axis, such as ^5 of Fig. 73, 
undergoes a helical deformation AD, and each elemental area 



96 



THE ELEMENTS OF MECHANICS OF MATERIALS 



of a section is subjected to a shearing stress produced by a strain 
called torsion, the stress varying directly as the distance of the 
elemental area from the axis. The angle 6 through which a 
longitudinal element is twisted is called the angle of shear , and 
varies directly as the distance of the element from the axis. 
The angle a is called the angle of twist, and is proportional to 
the length of the shaft. 




Fig- 73. 



Suppose a section of length dx, Fig. 74, to be cut from a shaft 
by planes perpendicular to the axis, and let da be the angle of 

twist for this section. The angle of 
twist varying with the length of the 
shaft, we shall have 




dx 



da 



whence da = 



idx 



Fig. 74. 

Expressing 6 and da in radians, we have 

BD = 6dx = cda, whence 6 = —— = — j 

dx L 



c being the radius of the shaft. 
From Art. 30, p. 62, we have 

Unit stress 



Unit deformation d 



A=G, 



whence 



S=Gd 



Gca 



S being the unit stress at the outermost surface. 



bHAFTS 



97 



Suppose the section of the shaft, Fig. 75, to be made up of a 
great number of small areas, a, ai, 02, etc., distant y, yi, 3/2, etc., 
from the axis. Each of these small areas is subjected to a shear- 
ing stress s, 5i, 52, etc. 




Fig. 75. 

If S denotes the unit stress of the material at the outermost 
fiber, distant c from the axis, it is the maximum unit stress to 
which the material will be subjected, since the stresses vary 
directly as the distance from the axis. 

The resisting moment of the section will be the sum of the 
moments of the elem^tal resistances, as, aiSi, a^s^, etc., and 
since there must be equilibrium between the internal and external 
forces, the resisting moment will equal the twisting moment and 
we shall have 

Mt = asy + ^1^13^1 + (hS2y2 + etc. 
But o 



Then 



M, 



c 



whence s=~- 



Si = 



Syi 



S2 



Sy2 
c 



Saiy^ S(hyt 



+ 



+ etc., 



= - (ay^ + aiyi^ + 02 V + etc.) = — 2, 
c c 

in which I^ is the polar moment of inertia. 



98 THE ELEMENTS OF MECHANICS OF MATERIALS 

Then s^^ = ^, 

whence a = -—— radians. 

Crip 

For a circular section /_ = Hence 

A 1 r^ •^- J 32^^/^X57.3 SM^tL 

Ande of twist m degrees = ^-^-^ = ^-^ — ^— , 

in which D is the diameter of the shaft. 

For a hollow circular shaft of external diameter D and internal 

diameter d, I^ = — ' and 

32 

Angle of twist for hollow shaft = ^_^ — ^— • 

SI 

From the equation Mt = — -^ we get 

c 

in which Zp denotes the polar modulus of the shaft section. 
It is thus seen that the strength of a shaft to resist torsion varies 
as its polar section modulus, just as the strength of a beam to 
resist bending varies as its plane section modulus (see Art. 34, 
p. 66). 

From what has preceded it is known that the least plane 
moment of inertia, /, of a circle is just one-half its polar moment 
of inertia, /p. That /p should be greater than / might have 
been expected, since the material of a circular shaft is in far 
better form to resist torsion than to resist bending. 

For shafts of small diameter and comparatively long lengths, 
such as line shafting in mills and shops, the angle of twist may 
be too great to insure safe transmission. In such cases the diam- 
eter of the shaft should be determined with reference to the 



SHAFTS 99 

Stiffness of the material rather than to its strength, the deter- 
mining factor being the angle of twist. 

The angle of twist should enter into the design of shafts under 
ten inches in diameter, and it may be accepted as a result of 
experience that if the angle of twist does not exceed i° in i8 diam- 
eters of length, shafts are sufficiently stiff. 

/^ 

We have shown that 5 = — - , in which a. is expressed in radians, 

and c = — • With the limiting value of L = i8 Z> for i° of 

2 

twist, we shall have 



57.3 X 18 L> 2063 

Taking Gas]io,5oo,ooo for wrought iron and as 12,500,000 for 
steel, we have for light shafting: 

For wrought iron, S = 5000 pounds per square inch. 
For steel, S = 6000 pounds per square inch. 

Example. — A steel shaft 2 inches in diameter and 10 feet 
long is subjected to a twisting moment of 7500 pounds-inches. 
What is the angle of twist? Is the shaft safe? 

Solution. — 

Angle of twist = ^-^^ = 584X7500X10X12 ^ 
^ GD^ 12,500,000 X 16 ^ 

An 1,1 1 r ^ • -L Length 10X12 o 

Allowable angle of twist = ^ ,. = = ^.S • 

^ 18 diameters 18 X 2 '^ '^ 

The shaft is therefore safe. 

PROBLEMS 

1. A 3-iiich steel shaft, 200 feet in length, is subjected to a twisting 
moment of 35,000 pounds-inches. Is it safe? Ans. Unsafe. 

2. If the shaft of problem i is unsafe, what should be its diameter to 
insure safety? Ans. 3.07 ins. 



lOO THE ELEMENTS OF MECHANICS OF MATERIALS 

47. Combined Torsion and Bending. — In cases of heavy- 
shafting, such as propeller shafts of vessels, the weight of the 
shaft occasions a bending action which increases the stress 
in the material over that due to torsion alone. The effect 
of this combined action of torsion and bending is not exactly 
determinable in practice, owing to the inexact computation of 
the bending moment involved. It may be shown, however, 
that if AI and Mt denote the bending and twisting moments 
respectively at any section of a shaft, and Me the equivalent 
twisting moment that would produce the same intensity of 
stress as that due to the combined torsion and bending, we 
shall have 



The mean t^^sting moment of shafts of steam engines is 
determined from the mean pressure in the cylinder, and the 
greater the ratio of expansion of the steam the greater difference 
there will be between the maximum and minimum moments to 
which the shaft wiU be subjected; and since a shaft must be strong 
enough to resist the maximum stress to which it may be liable, 
the twisting moment expressing the maximum stress due to the 
combined twisting and bending should be the basis of the cal- 
culation to determine its diameter. For practical purposes the 
equivalent twisting moment may be taken as 1.5 times the 
mean twisting moment, and in some exceptional cases this 
multipHer may be increased to 2. 

48. Transmission of Power by Shafts. — If P denotes the 
mean force acting on the crank pin of an engine at a distance of 
r inches from the axis of the shaft, then Fr is the mean twisting 
moment of the shaft in pounds-inches. 

For N revolutions of the shaft per minute the path of P is 

2 irrNP 
2 irrN inches, and the work performed is foot poimds. 



SHAFTS 10 I 

Then 

Horse power transmitted = H.P. = , 

12X33,000 

■ „ , , 12 X 33,00 X H.P. 63,025 X H.P. 
whence Pr = Mt= ^^ — = -^ — ^ • 

We have shown that 

Mt = — - = = 0.196 5Z)^ 

c 16 

in which D is the diameter of the shaft. 

For short and heavy shafts where the angle of twist is not 
important, the value of S may be taken as 10,000 pounds per 
square inch for steel and as 8000 for wrought iron. Taking the 
equivalent twisting moment as 1.5 Mty we shall have for heavy 
steel shafts: 

^ y63,o25 X H.pixT^ ^ ygp. , 

V 0.196 X 10,000 N ^^^V N ' 
and for heavy wrought-iron shafting, where S = 8000, 

The light shafts of machine and wood-working shops, rimning 
at from 125 to 250 revolutions per minute, and carrying pulleys 
from which machines are (Jnven, are subjected to bending as well 
as torsion. In such cases it is in accordance with good practice 
to take the equivalent twisting moment at 1.5 times the mean, 
and to take the value of S for steel as 6000 pounds per square 
inch, giving for the diameter the value 



^ = 4.3^^ 



N 

For hollow shafts in which the internal diameter d is one-half 
the external diameter D, the common practice for shafts of 
steamships, we have 

c 32 c 



I02 THE ELEMENTS OF MECHANICS OF MATERIALS 

Therefore, for hollow shafts of steel, with 5 = 10,000 and the 
equivalent twisting moment 1.5 times the mean, we have 



^ 7 63,0.5 X H.P. X 1.5 ^ ^ILP. . 

V 0.184 X 10,000 iv ^ ' y N 

For hollow iron shafts, S = 8000, and 



Vh.] 



D = W^- 



Example. — A steel shaft is to have 30 per cent of its section 
area removed in making an axial hole, and is to transmit 9000 H.P. 
at 120 revolutions per minute. Taking the equivalent twisting 
moment as 1.5 times the mean twisting moment, find the internal 
and external diameters of the shaft; show that the shaft is 25.55 
per cent lighter than a solid shaft would be to transmit the same 
power; and show that the two shafts would be of equal strength. 

Solution. — Area of removed section = '^^ — . 

4 

Therefore — = -^ , and ^ = 0.5477 Z). 

4 4 

2TrNPr 



H.P. = 



12 X 33,000 



whence 

„ ,^ 9000X12X33,000X1.5 J . -, 

i.K Pr=Mt = "^^ = 7,090,000 pounds-mches. 

2 ttX 120 

c 2>^c 16D 

Then 1786.78 Z)^ = 7,090,000, 

whence Z) = 15.83 inches, 

and iZ = 0.5477 Z) = 8.67 inches. 

A solid steel shaft to transmit 9000 H.P. at 120 revolutions per 
minute would have a diameter 



D = 3.64 y ^^^ = 15.35 inches. 



SHAFTS 103 

Since the weights of shafts are proportional to their sectional 
areas, and as the areas vary as the squares of their diameters, 
we find the hollow shaft to be 

1001(15.35)2 - [(15.83)2 - (8.67)2]} 

^^ ^^^' Lv ^ ^; ^ — i_n± ^ 25.55 per cent 

(i5-35r 

lighter than the soHd shaft. 
For the hollow shaft 

Z, = ^ = ^ = 0.1787 D' = 0.1787 X (15.83)^ = 709. 

For the solid shaft 

^ ttD^ ttD^ 3.1416(15.35)3 

Z^ = = -— = ^-^ — \ ^ ^^^ = 709, 

32c 16 16 

The moduH of the sections being equal, the shafts are of equal 
strength. 



PROBLEMS 

1. Find the diameter of a wrought-iron shaft to transmit 90 H.P. at 
130 revolutions per minute. What should be the diameter if there is a 
bending moment equal to the twisting moment ? 

Ans, 3.54 ins.; 4.75 ins. 

2. Find the diameter of a» steel shaft for a steam engine having an 
overhung crank. Diameter of cylinder, 18 inches; mean steam pressure, 
130 pounds per square inch; stroke of piston, 30 inches; overhang of 
crank, i.e. middle of crank pin to middle of bearing, 18 inches. Use 
Me=M + VM2 + Mt^. Ans. 9.56 ins. 

3. A 4-inch steel shaft 30 feet long is found to have an angle of twist 
of 6.2° when transmitting power while making 130 revolutions per minute. 
Find the H.P. transmitted. Ans. 194.64. 

4. Find the diameter of a steel shaft, 50 feet long to transmit 30 H.P. 
at 210 revolutions a minute. Ans. 3.2 ins. 

5. Find the diameter of a hollow steel shaft to transmit 10,000 H.P. at 
120 revolutions per minute, the external diameter being twice the inner, 
and the equivalent twisting moment 1.5 times the mean. 

Ans. 16.25 ins. 



I04 THE ELEMENTS OF MECHANICS OF MATERIALS 

6. A steel shaft is to have 25 per cent of its sectional area removed in 
making an axial hole, and is then to transmit 7200 H.P. at 120 revolutions 
per minute. Taking the equivalent twisting moment as 1.5 times the 
mean, find the internal and external diameters of the shaft, and show that 
the shaft is 21.7 per cent Hghter than a soHd shaft would be to transmit the 
same power. Ans, 14.56 ins, and 7.28 ins. 

7. The main shaft of a machine shop, in transmitting 40 H.P. at 120 
revolutions per minute, carries the average number of pulleys. Find its 
diameter if made of steel. Ans. 3 ins. 



CHAPTER VI 

INTERNAL WORK DUE TO DEFORMATION. SUDDENLY 
APPLIED LOADS 

49. Resilience. — The stresses thus far encountered have 
been those produced by external forces gradually applied to 
bodies under static conditions, and the resulting deformations 
necessarily produced internal work in the bodies. As the stresses 
produced were within the elastic limit, the bodies resumed their 
original shapes upon the removal of the acting forces, and in 
doing so the internal work due to the deformations was, in 
accordance with the principle of the conservation of energy, 
given out in the form of mechanical energy. It is thus seen 
that the internal work of deformation is a form of potential 
energy, called resilience. 

50. Effect of Suddenly Applied Loads. — The sudden appli- 
cation of loads, such as a train passing rapidly over a bridge, 
causes greater deflections than would be the case were the same 
loads applied gradually, ^hese deflections are but momentary, 
as resiHence causes a vibratory action in the affected bodies 
until the effect of the shock due to the sudden application of the 
load disappears. 

If a load W be gradually applied to a beam, producing a stress 
S, the intensities of the load and stress will gradually increase 
from o to W and from o to 5 respectively, the mean values 

TTT- O 

being — and -. If ^^ be the deflection, we shall have for the 
2 2 

equality of the external and internal work 

W S 

— Xy = - Xy, whence S = W, 

2 2 



I06 THE ELEMENTS OF MECHANICS OF MATERIALS 

If the same load be applied suddenly, its intensity remaining 
constant during the period of action, the resulting stress S' 

will increase from o to S\ its mean value being — , and a deflec- 

2 

tion y' will be produced. We shall then have 

Wy =- X y, whence S' = 2W. 
2 

In other words, the sudden application of a load produces a 
stress twice as great as that produced by the same load applied 
gradually. 

Should a load W fall on a beam through a distance h and pro- 
duce a deflection y, the kinetic energy developed would be 
W^(h + y) ; but all this energy would not be converted into work 
of deformation on account of the inevitable loss of kinetic energy 
sustained by all partly elastic bodies during impact. 

Example I. — What load may fall through a distance of 
8 inches on the middle of a 12-inch Cambria I beam in order that 
the maximum fiber stress shall not exceed 20,000 pounds per 
square inch, supposing 70 per cent of the kinetic energy of the 
falling load to be converted into work of deformation? The 
length of the beam is 16 feet, and its tabulated moment of 
inertia is 215.8. 

Solution. — Let W denote the gradually applied load that 
would perform, in producing a deflection y, the same work of 
deformation as that performed by the falling load W\ The 

W . . . Wy 

mean value of PT is — , and its work of deformation is — -- 

2 , 2 

WL 

The bending moment at the middle section due to W is f 

4 
and we shall have 

e, Mc WLc , „, ^SI 

S = —r- = ;:- , whcUCC W = ^=V~ * 

74/ Lc 

^ WD 

The deflection due to W we have found to be 



48£/' 

(Art. 39, Ex. I). 



INTERNAL WORK — SUDDENLY APPLIED LOADS 107 

WD SD 20,000(16X12)2 . , 

Hence y = -—=rp = — tt = — ^ 777 = o-34i inch. 

48 A/ 12 he 12 X 30,000,000 X 6 

Then 

Work of deformation by W = — - = —1 — X =r = t~T7^ 

2 Lc 12 Ec 6 Ec^ 

(20,000)2 X 215.8 X 16 X 12 ^ z- • u J 

= ^ — — — ;: = 2557.0 inch-pounds. 

6 X 30,000,000 X 36 

Work of deformation by W' is 0.7 W^ (8 + 0.341), hence 
2557-6 = 0.71^^X8.341, 
whence W'= 438 pounds. 

PROBLEM 

I. From what height may a load of 1200 pounds fall on a 15-inch 
Cambria I beam 15 feet long and having a moment of inertia of 511, in 
order that the maximum fiber stress shall not exceed 30,000 pounds per 
square inch, and supposing 70 per cent of the kinetic energy of the falling 
weight to be converted into work of deformation ? 

Ans, 9.37 ins. 

The sudden application of loads produces impact, resulting in 
stresses greatly in excess of those produced by the same loads 
when applied gradually, and in the recovery from the shocks 
thus occasioned the material sustains such rapid vibrations that 
its molecular structure and elasticity may be impaired. In 
the design of machines and structures that are subjected to 
shocks the effect of impact is an important consideration, and the 
determination of the resilience of a given material is the best 
measure of its ability to withstand shocks. 

51. Modulus of Resilience. — If a bar be placed in a testing 
machine and subjected to a gradually increasing load in a direc- 
tion to produce a tensile stress, the elongation produced will be 
proportional to the load, provided the elastic limit of the material 
be not exceeded. The load, or external force, will gradually 

increase from o to W, and its mean value will be — li y 

2 



I08 THE ELEMENTS OF MECHANICS OF MATERIALS 

denotes the elongation, the expression for the work done is — -- 

2 

It is a fundamental assumption that the total stress in the bar 
is uniformly distributed throughout its section, so that if A 
denotes the area of a section of the bar and S denotes the unit 
stress, we shall have W = AS\ and it has been shown in Art. 31, 

SL 

p. 63, that y = -=r, so that the expression for the work in this 
hj 

instance may be written 

Internal work = — - • AL. 
2E 

But A Lis, the volume of the bar, consequently the work done is 

proportional to the volume of the bar, and therefore to its weight. 

The work done in stretching the bar to its elastic limit is its 

resilience, and if S be taken as the unit stress at the elastic limit, 

. 5^ . 
the ratio — =, is known as the modulus of resilience. The resilience 
2E •' 

of the bar is then found by multiplying its volume by the modulus 

of resilience. 

Example II. — A steel bar 10 feet long and 2 inches in diameter 
is stretched to the elastic limit; what is its resilience? 

Solution. — The elastic limit of steel in tension is 50,000 pounds 
per square inch, and the coefficient of elasticity is 30,000,000 
pounds per square inch. 

rr^i lijT 1 1 r '^• -S^ (SO^OOO)^ 2^0 

Then, Modulus of resilience = — - = ^^^-^ = -^ ? 

2E 2 X 30,000,000 6 

and Volume of bar = 3.1416 X 120= 377 cubic inches. 

Then, Resilience of bar = -^ — ^'' = 15,708 inch-pounds. 

PROBLEMS 

I. In gradually elongating a steel bar i square inch in section and 20 
feet long, work to the amount of 150 foot pounds is expended. Find the 
applied force and the amount of elongation produced. 

Ans. 21,213 lbs.; 0.17 in. 



RESILIENCE OF BEAMS 109 

2. A wrought iron bar 4 inches in diameter and 10 feet long has its 
unit tensile stress increased from 10,000 to 20,000 pounds per square 
inch. What additional potential energy is stored in the rod by the oper- 
ation? Ans. 673 ft. lbs. 

3. If 600 foot pounds of work are expended in increasing the unit 
tensile stress of a steel bar 4 inches in diameter from 10,000 to 20,000 
pounds per square inch what is the length of the rod? Ans. 7.5 ft. 

52. Resilience of Beams. — The work performed in bending 
a beam to the maximum deflection within the elastic limit is 
the resilience of the beam. Denoting the load by W, the stress 

in the beam increases from o to W, the mean value being — 

2 

The work performed, or the resilience, is therefore equal to half 
the load multiplied by the deflection. Several illustrations of 
the methods of finding the resilience of beams under different 
conditions of loading will here be given. 

I. Cantilever with load at extremity. — The maximum deflection 

bemg — — (Art. 39, Ex. II), the resdience is -— - . — = — — - • 
3 iii 3 iii 2 lii 

SI SI 

Mmsix = WL = — , whence W = -— - 
c Lc 

S'^PU S'^IL S'^ K^ 

Then, Resilience = = —-, ; = -—,' — 'AL, 

6 EIUc^ 2 £ . 3 c^ 2 E 2>c 

in which AK^ is substituted for /. 

For rectangular sections, c = - and K} = — = — — = — , so 

2 A \2hd ■ \2 

K^ I 
that ^r = - . Then, for cantilevers of rectangular section and 

c" 3 
with concentrated load at the free end, we have 

Resilience = — = , 

2 E 9 

or it is the product of the modulus of resilience and one-ninth 
the volume of the beam. 



no THE ELEMENTS OF MECHANICS OF MATERIALS 

II. Simple beam with load concentrated at middle. — The 

WD 
maximum deflection being ——-(Art. 39, Ex. I), the resihence 

40 hi 

. WD W _ W^D 
^^ 48 £/ * 2 ~ 96 EI ' 

,, WL SI , „, 4 SI 

Mrnax = = — , whence W = =V~ • 

4 c Lc 

rr.. J. -r 16 S'PD S'^IL , S' K^ ,^ 

Then, Resihence = ^^^^, = ^^-^, = ^'-.' ^L. 

K^ I 

For rectangular sections, -r- = ~ ? and we shall have for a simple 

c' 3 
beam with load at middle 

^ .,. S' AL 

Resilience = — - • — , 
2 h 9 

which is the same as that found for a cantilever with load at its 
extremity. 

III. Cantilever uniformly loaded. — If a cantilever be loaded 
with w pounds per linear unit, the elemental load is wdx^ and 
if y be the corresponding deflection the elemental external work 

uwvdoc 
is -^ — , the summation of which will be the resihence. 

2 

In Art. 39, Ex. IV, we found 

y = -^ (6 Dx' -4Lx' + x'). 
24 hi 

w^ C^ 
Then, Resilience = -;— ^ 1 (6 LV — 4 Lx^ + ^^) dx 
48 EI Jq 



Udx^- Lx' + ^ - 



6w^D 



48 £/ L 5 Jo 240 EI 

The bending moment at any section distant x from the wall 

w (L — xY w (D — 2 Lx -\- x^) , . 

IS ^ = ^ , and IS a maximum 

2 2 

when ii; = o, hence 

,, wD SI , 2 SI 

Mmax = = — , whence w = — - • 

2 c Dc 



INTERNAL WORK — SUDDENLY APPLIED LOADS III 

Then, Resilience = — - •-.—-• ylZ. 

2E s c^ 



Resilience 



For rectangular sections, -^ = - , hence 

3 

52 AL 



c' 3 



2E 15 
for cantilevers of rectangular section and uniform load. 

IV. Simple beam uniformly loaded. — ■ The elemental load is 
wdx, and if y be the corresponding deflection, the elemental 

iwdoc 
external work is — — , the summation of which will be the 

2 

resilience. 

In Art. 39, Ex. Ill, we found 

y = =- (2 Lx^ — x^ — Dx), therefore 

24 hi 

Resilience = / (2 Lx^ — x^ — Ux) dx 

48 hi J Q 



VLx^ _ ^ _ L^~f 
L 2 5 2 Jo ~ 



w^L^ 



48 £/ L 2 5 2 Jo 240 EI 

The bending moment at any section distant x from the left 

ivLx "wx iv 
support is = — {Lx — x^), and is a maximum at the 

222, 

middle of the beam, where x = - , hence 

2 

,. wD SI , SSI 

Mma^ = —— = — , whence w = -—- • 
8 c Uc 

Then 

^•2 g ^2 
ResiHence = — * AL. 

2E 15 r 

K^ I 

For rectangular sections, ^7 = - , hence 

c" 3 

^ .,. 52 SAL 

Resihence = — - • 

2^ 45 

for beams of rectangular section and uniformly loaded. 



112 



THE ELEMENTS OF MECHANICS OF MATERIALS 



53. Resilience of Torsion. — Suppose the shaft of Fig. 76 to 
be subjected to a twisting moment P'r' , and in consequence 

■p' suppose 3; to be the 
displacement of an ele- 
mental area a distant 
r from the axis. If R 
denotes the unit shear- 
ings tress in area a, 
then 

o -\- Ra ^^ Ray 

Xy = — -' 

2 -^ 2 




Fig. 76. 

Work performed in displacing a = 



From Art. 31 we shall have 

G R . RL 

- = — J whence y = -—- 
L y G 



Then. 



Work performed in displacing a 



R^aL 
2G 



Total work performed = — — - H — | — [-etc., 

2 G 2 G 2 G 

which is an expression for the resilience of torsion of the beam. 

If S denotes the unit shearing stress in the remotest ele- 

S . 
mental area, distant c from the axis, then - is the unit shearing 

c 

Sr 
stress at a unit's distance from the axis, and — is the unit shear- 

c 

ing stress in the elemental area a distant r from the axis. 
Hence R = — > and in Hke manner Ri = — -^ and R2 = — -- 



We shall then have 

Resilience of torsion = — ^7^ H ^;-^ — 1 ;;-;; — h etc. 



2Gc^ 



2Gc^ 
S'LI 



2 Go' 2 Gc^ 

{ar'^ -f airi^ + ^2^2^ + etc.) 



2 Gc^ 2 G c^ 



2G r 



since the polar moment of inertia Ip = AK^. 



RESILIENCE OF TORSION 1 13 

For circular sections, c = — and K^ = ~ = r = — > 

2 ^ 32 4 8 

so that -T = ~ • 

c-^ 2 

Then, for shafts of circular section we have 

S"^ AL 

Resilience of torsion = — — > 

2G 2 

or, it is the product of the coefficient of resilience for torsion and 
one-half the volume of the shaft. 

Comparing the expressions derived for the resiHence of a bar 
in tension, of beams of rectangular sections under different con- 
ditions of loading, and of a circular shaft under torsion, we ob- 
serve that the resilience of the bar is 9 times as great as that of 
a cantilever loaded at its free end, 9 times that of a simple beam 
loaded at the middle, 15 times that of a cantilever uniformly 
loaded, 5f times that of a simple beam uniformly loaded, and 
twice that of a shaft under torsion. 

PROBLEMS 

1. Derive the expression for the resilience of a beam uniformly loaded 
with w poimds per linear unit and fixed at the ends. . S^ AL 

2E' 15' 

2. Derive the expression Tor the resilience of a beam fixed at the ends 

and having a load W concentrated at the middle. . S^ AL 

Ans. —=. 

2E Q 



CHAPTER VII 

GRAPHIC STATICS 

System of Lettering. Force Diagram. Funicular 
Polygon 

54. Graphic Statics. — The methods by which static problems 
are solved by means of scale drawings constitute graphic statics. 

In very many cases the determination by calculation of the 
forces transmitted through the different parts of a structure 
involve tedious and difficult processes, with the consequent 
liability to error, and in the end the effort to check the accuracy 
of the results occasions a procedure as prolonged and tedious 
as the processes themselves. By the graphic method, however^ 
solutions are readily obtained, and the process itself furnishes a 
check as to accuracy. 

55. System of Lettering. — To the system of lettering dia- 
grams devised by A. H. Bow is due much of the facility in 
making graphic solutions. 

The two important features of the Bow system of lettering 
are: (i) The placing of a letter in each of the spaces between 
the lines of action of the external forces; (2) the naming in clock- 
wise order of each force by the two letters flanking it. 

The forces of Fig. 77 will serve to illustrate the Bow system. 
The five forces are in equilibrium, and the letters A, B, C, D, 
and E are placed in the spaces between them. Other letters, 
or even numbers, would serve the purpose, and they might be 
placed in any order, but it will be found convenient to commence 
at the left and letter the spaces alphabetically. The force of 
20 pounds is flanked by the letters A and B and is known as the 

114 



GRAPHIC STATICS 



115 



force AB, not as BA, since the letters must be read clockwise. 
In like manner the remaining forces are known as BC, CD, DE, 
and EA. 



' V 5 lbs. 



y 15 lbs. 



Fig. 77. 



h 

Fig. 78. 



Should the system of forces act on an open-framed structure, 
such as a roof truss, Fig. 79, a letter must be placed within each 
open space of the frame in addition to those placed in the spaces 
between the external forces. 
The external forces, by the Bow 
notation, are known as AB, BC, 
CD, DE, and EA. Th& stress 
in the member connecting joints 
I and 2 is known as BG. The 
stress in the member separating 
the spaces lettered F and G, may ^^^- '^^• 

be known as GF or as FG, according to the end of the member 
under consideration. If the end considered is that at joint i 
then the stress in the member is known as GF, because the forces 
about that joint, taken in clockwise order, are AB, BG, GF, and 
FA . If the other end of the member is under consideration the 
stress is known as FG, because the forces about the joint at that 
end, in clockwise order, are known as FG, GH, HI, IE, and EF. 




Il6 THE ELEMENTS OF MECHANICS OF MATERIALS 

56. Force Diagram. — Taking the forces of Fig. 77 in clock- 
wise order, and denoting them by the corresponding small let- 
ters of the alphabet, we may represent them in magnitude and 
direction in a diagram by lines drawn to some chosen scale. 
Such a diagram is known as deforce diagram. Thus, A being the 
first letter in clockwise order from the left, the letter a will be 
the starting point of the force diagram of Fig. 78, and since the 
force AB acts downward, ah, one inch in length, will represent 
it to the scale of 20 pounds to the inch. The force BC, next in 
clockwise order and equal to 40 pounds, acts upward and will be 
denoted by he, 2 inches in length and measured upward. In like 
manner, cd measured downward and one-half inch in length, de- 
notes the force CD of 10 pounds ; de measured upward and one- 
quarter inch in length, denotes the force DE of 5 pounds; and, 
finally, ea, which is found to measure three-quarters of an inch, 
properly denotes the force EA of 15 pounds. It will be noted that 
the force EA is of just sufficient magnitude to fill, when drawn to 
scale, the space between e and a and thus close the force diagram. 
This force diagram is, in reality, a closed polygon which has 
resolved itself into a straight line in consequence of all the forces 
being parallel. Had it not closed, the system being in equilib- 
rium, there would have been an error in the construction. 

With a correct construction, and the last point of a force 
diagram not falling on the first point, the construction would 
indicate a resultant force, equal in magnitude to the scale dis- 
tance between the last point and the first point, and acting 
upward or downward according as the last point had fallen above 
or below the first point. 

If one of the forces of Fig. 77 were unknown it could be found 
by means of the force diagram. Suppose the force BC unknown. 
Then, commencing with the force CD, the force diagram would 
be constructed by representing in clockwise order the forces CD, 
DE, EA, and ^^ by the scale distances cd, de, ea, and a^respec- 



FUNICULAR POLYGON II 7 

tively, of Fig. 78, c being the first point of the diagram and h 
the last thus determined. The missing force must then be repre- 
sented by hc^ measured upward, and as it measures 2 inches, it 
correctly does so for the force BC of 40 pounds. 

The system of lettering enables the resultant of any number 
of the forces to be read at once from the force diagram. Sup- 
pose the resultant of the forces 5C, CZ), and BE were required. 
The first and last letter in the naming of these forces are B and 
E respectively, so that he on the force diagram represents the 
required resultant in magnitude and direction. The measure- 
ment of he is found to be 1.75 inches, which, to the scale, repre- 
sents 35 pounds, the resultant of the forces 40 -f 5 — 10, and it 
acts upward, since he is measured upward. 

57. Funicular Polygon. — Suppose a jointed frame, Fig. 80, 
to be acted on at its hinged joints by a system of forces in equi- 
librium, the members, bars, or links of the 
frame being free to adjust themselves to 
the best position for withstanding the 
action of the forces. Such a figure is 
called a funicular polygon, the word funic- 
ular having no mechanical significance. 

Since the system is in equilibrium, the 
funicular polygon must, of course, be a 

Fig. 80. 

closed polygon. The equilibrium is occa- 
sioned by the balance between the internal forces, or stresses, 
in the members and the external forces, and the whole being 
in equihbrium, each joint in itself is in equilibrium. 

Since the equilibrium at each joint is the result of the action 
of three concurring forces, viz., the external force at the joint 
and the stresses set up in the two members, it follows that a 
triangle may be constructed for each joint which will represent 
these forces in magnitude and direction. 

If, for example, the force ABht known, the equilibrium of the 





Il8 THE ELEMENTS OF MECHANICS OF MATERIALS 

joint at which it is applied is maintained by the action of the 
external force AB and the stress forces in the members BO and 
OA. Draw ab, Fig. 8i, parallel to the force AB, and make its 
length, to some chosen scale, represent the 
magnitude of ^-B. From b and a draw Hnes 
parallel to BO and OA respectively, inter- 
secting at 0. Then abo is the triangle of 
forces for the joint ABO, and bo and oa 
represent not only the directions of the 
stresses in the members BO and OA, but 
their magnitudes as well, to the same scale as ab represents 
the force AB. The stress bo in the end of the member 
BO at which AB is applied occasions an equal and opposite 
stress ob at the other end of BO, and, in fact, such is the 
case in all of the members of the polygon, for in no other 
way could the equilibrium be maintained. Taking the next 
joint in clockwise order, we have the external force BC and 
the stress forces CO and OB in equihbrium. But ob has 
just been found to represent in magnitude and direction the 
stress OB. Hence, by drawing from b and o lines parallel re- 
spectively to BC and CO, intersecting at c, we shall have bco 
as the triangle of forces for the joint BCO, and be and co will 
represent in magnitude and direction the force BC and the stress 
in the member CO respectively. We now know the stress oc at 
the joint CDO, and can construct the triangle of forces, cdo, 
for that joint. In like manner we can proceed and determine 
all the external forces and stresses in the members, the last line, 
fa, closing the diagram, thus proving that the system is in 
equihbrium. 

It will be observed that the sides of the polygon just con- 
structed represent the external forces of the funicular polygon, 
and therefore abcdef is the force diagram. Furthermore, all the 
lines representing the stresses in the members of the funicular 



FUNICULAR POLYGON II9 

polygon meet at a point called the pole. These stress Hnes are 
known as vectors. 

From what has preceded it is seen that if all the external 
forces that are applied to a funicular polygon are known in 
magnitude and direction, and also the directions of two of its 
members, the force polygon can be drawn. For the force 
diagram can be drawn from the known forces, and the inter- 
section of the vectors parallel to the known directions of the two 
members gives the pole. The directions of the remaining mem- 
bers are then found by drawing the rest of the vectors. 

A force diagram of any system of forces in equilibrium can 
be drawn, and by choosing any pole, a funicular polygon with 
respect to that pole can then be constructed to which the forces 
may be applied. 

If the magnitudes and directions of a system of forces acting 
on a body are known and the system is not in equilibrium, the line 
of action of the force required for equilibrium may be determined 
by means of the funicular polygon. For the force diagram of 
the given forces may be drawn and the gap representing its lack 
of closure will, by the polygon of forces, give the magnitude and 
direction of the resultant. A pole for this force diagram may 
be selected arbitrarily anjd vectors drawn. Starting at a selected 
point in the line of action of any one of the forces, a funicular 
polygon may be drawn with respect to the chosen pole, and the 
intersection of the two members of this funicular that are parallel 
to the vectors drawn to the extremities of the resultant in the 
force diagram gives the joint of the funicular at which the re- 
quired force, equal and opposite to the resultant, must be applied. 

PROBLEMS 

I. The parallel forces of Fig. a .^ ^^ 

are in equilibrium. Find by a force I 
diagram the magnitude of the force 



E 

BC. Ans, 17. Yig. a. 



I20 THE ELEMENTS OF MECHANICS OF MATERIALS 



2. Forces of 20 pounds, 25 pounds, 15 pounds, 
and 30 pounds act on a body in the directions 
shown in Fig. h. Find the magnitude and direc- 
tion of the resultant. Find also by means of 
a funicular polygon the line of action of the 
force required for equilibrium. 

Ans. Resultant = 11 lbs. 



Fig- b' 58. Illustrations. — To illustrate prop- 

erties of the funicular polygon the closed polygon of Fig. 82 
has been drawn, having the known forces AB, BC, CD, DE, and 
EA acting at its joints. 






Fig. 82. 



Fig. 83. 



Draw to a selected scale the force diagram abcde of Fig. 83. 
The triangle of forces abo for the joint ABO determines the 
pole 0, and the other vectors may then be drawn. 

Let a plane xy divide the polygon into two parts. The forces 
of the part to the right of xy are AB and BC, and, by the triangle 
of forces, their resultant is ac, Fig. 83, acting from a to c. The 
forces of the part to the left of xy are CD, DE, and EA, and, by 
the polygon of forces, their resultant is ca, acting from c to a. 
Hence, the resultant of the forces of one part has the same mag- 
nitude and Hne of action as the resultant of the forces of the 



FUNICULAR POLYGON 121 

other part, but acting in opposite directions, showing that the 
resultant of the forces of one part maintains equihbrium with 
the forces of the other part. 

To find where the resultant of the two forces to the right of %y 
acts, we replace the forces ab and he in the force diagram by their 
resultant ac, so that our force diagram now becomes acde. The 
vectors oc, od, oe, and oa of the force diagram have o as their pole, 
so that a funicular polygon may be drawn with respect to o, 
having its sides parallel to these vectors. We already have OC, 
OD, OE, and OA of Fig. 82 parallel to these vectors, but they do 
not close the polygon, and since a funicular polygon must close 
we produce OC and OA until they intersect, and at the joint thus 
formed the resultant R, having the magnitude and direction 
of ac, will act as shown by the dotted lines of Fig. 82. By a 
similar process the resultant, ca, of the forces CD, DE, and EA 
may be shown to act at the same joint but in the opposite 
direction. 

An inspection of the funicular polygon of Fig. 82 and of the 
force diagram of Fig. 83 shows: 

{a) The resultant of the forces CD, DE, and EA to the left 
of the section xy is given by ca, the first and last letters of the 
forces when named in clgckwise order; in like manner the re- 
sultant of the forces AB and BC to the right of xy is ac. 

(b) The letters in the force diagram which name the resultant 
also name the members of the funicular which have to be pro- 
duced to their intersection in order to get a point in the line of 
action of the resultant. Thus, the resultant of the forces CD 
DE, and EA is ca, and by producing the members C and A to 
their intersection in Fig. 82 a point in the line of action of the 
resultant is obtained. 

It will be observed that the introduction of the external force 
R, the resultant of the forces AB and BC, changes the original 
funicular polygon to one having fewer joints by one. Should the 



122 THE ELEMENTS OF MECHANICS OF MATERIALS 

members C and E be produced to their point of intersection 
and the resultant ce of the forces CD and DE be applied 
at the point, the funicular would be reduced to one of three 
joints. 

The members cut by the section xy are A and C, and the 
stresses in these members, in magnitude and direction, are oa 
and oc of Fig. 83. On the right side of xy the stresses in the 
members A and C act in the directions shown by the arrowheads, 
and their resultant, in magnitude and direction, is ca, which is 
opposed by the equal and opposite external force R, or ac. On 
the left side of xy the stresses in the members A and C act in the 
directions indicated by the arrowheads, and their resultant is 
ac in magnitude and direction. The external forces to the left 
of xy are CD, DE, and EA, and their resultant in magnitude 
and direction is ca, which is opposed by the equal and opposite 
resultant ac of the stresses in the members A and C. It is thus 
seen that on either side of the section xy there is equilibrium 
between the external forces and the internal stresses in the mem- 
bers cut by xy, and on this principle is founded the section 
method of determining stresses, to be referred to later. 

A practical application of the funicular polygon will be made 
by taking the beam of Fig. 27, p. 31, reproduced in Fig. 84. 
The linear scale is i inch = 4 feet, and the load scale i inch = 
150 pounds. Then, Wi = 60 pounds = 0.4 inch to scale, 
W2 = 45 pounds = 0.3 inch, and W3 = 90 pounds = 0.6 inch. 
Letter the beam according to the Bow system, so that Wi, W2, 
TF3, R2, and Ri will be known as AB, BC, CD, DE, and EA 
respectively. 

To construct the force diagram we set off, vertically downward, 
ab equal in length to 0.4 inch to represent the downward force Wi, 
or AB, to scale; be equal in length to 0.3 inch to represent W2, 
or BC; and cd equal in length to 0.6 inch to represent W3, or 
CD. Then we know that da is the closing line of the force 



FUNICULAR POLYGON 



123 



diagram, all the forces being vertical, and that it represents 
the sum of the reactions Ri and R2; but we do not know the 
amount of the load borne by each support. 




Fig. 84. 



To find Ri and R2 we must construct the funicular polygon 
of the forces. Select at random some point 0, distant oh from 
the load line ab, as a pole and draw the vectors oa, ob, oc, and od. 
From some point j in the line of action of Ri draw a line parallel 
to the vector oa, and from its point of intersection, k, with the 
vertical from Wi draw a line parallel to the vector ob, and from 
its point of intersection, /, with the line of action of W2 draw 
a line parallel to vector oc, and from its point of intersection, m, 



124 THE ELEMENTS OF MECHANICS OF MATERIALS 

with the Hne of action of Wz draw a line parallel to the vector 
od. This last line intersects the line of action of R2 at n. Join 
n with j, and we have the funicular polygon jkfmn with the 
external forces of the beam acting at its joints. This funicular 
polygon has five sides, while there are but four vectors in the 
force polygon, and since there must be a vector for each side 
of the funicular polygon, the vector oe, parallel to nj, must be 
drawn. The point e is thus determined, and de represents in 
magnitude and direction the support reaction DE = R^ to the 
scale adopted; and in like manner ea represents in magnitude 
and direction the support reaction EA = Ri. By measurement 
de is H inch, which, reduced to scale, equals H X 150 = 127.5 
pounds = R2; and ea measures f^ inch, which, reduced to scale, 
is 67.5 pounds = Ri. These are the same values found before 
for Ri and R2. 

It will not be necessary to letter the funicular polygon in order 
to know the names of the members, as an examination of the 
force polygon at once discloses them. Thus, the member 
parallel to oa is OA, the one parallel to ob is OB, and so on. 
The names of the members may also be determined from the 
lettering of the beam, since the length of each member is termi- 
nated by the lines of action of two forces of the beam. For 
example, the member kf is terminated by the Hnes of action of 
Wi and W2, and since the letter B appears between Wi and W2 
the member kf is named OB, or simply B. 

The resultant, a^,of Wi, W2, and Wz acts through r (see Art. 57). 
Should it be desired to replace two or more of the forces by their 
resultant, its magnitude and a point in its line of appHcation 
may be determined. Thus, if it were desired to replace the 
forces Wz and R2 by their equivalent, we find from the force 
diagram that the resultant of cd and de is ce in magnitude and 
direction, and a point in the line of application is found by 
producing OE and OC to their intersection as shown. 



FUNICULAR POLYGON 12$ 

59. The Funicular Polygon a B ending-moment Diagram. — 

Consider any section a of the beam of Fig. 84. Produce jk to 
its intersection with the vertical from a at p. Draw the hori- 
zontals X and Xi and regard them as the altitudes of the triangles 
jpz and kpy respectively. All the triangles of the force polygon 
have the same altitude oh. 

The triangles jpz and kpy are respectively similar to the 
triangles oea and oha, having their sides mutually parallel. 
Hence we have 

—r = — =^ , whence Rix = pz X oh, 
oh ea Ri 

and -7 = ^ = ^ , whence WiXi = py X oh. 

oh ah Wi 

The bending moment at the section a is, 

Ma = Rix — WiXi = pzX oh — py X oh = oh {pz — py) =yzX oh. 

That is, the bending moment at any section of the beam is equal 
to the product of the ordinate of the funicular polygon at the 
section and the polar distance. Thus, the ordinate yz under the 
section a measures || inch, and the polar distance oh measures 
I inch, representing 150 pounds to the scale selected. Then the 
bending moment at section a is, 

Ma = fl X4X 150 = 217.5 pounds- feet, 

as was found on page 33. 

The shear diagram of the beam can readily be constructed 
by projection from the force diagram. Thus, the shear at any 
section between the left support and W\ is Ri = ea, and is 
plotted by projecting ea horizontally as shown. At any section 
between Wi and W2 the shear is Ri — Wi = ea ~ ab = eh, and 
is plotted by projecting eh. The shear at any section between 
Wi and TF3 is Ri — Wi — W2 = ea — ah — ho = — ec, and be- 
tween Wz and the right support the shear is Ri — Wi — W2 — W3 



12.6 THE ELEMENTS OF MECHANICS OF MATERIALS 

= ea — ah — be — cd = — ed = — R2. Projecting these two 
shears, the diagram is completed. 

It has been stated that the pole of the force diagram may be 
selected at random, but if it be selected so that its distance from 
the load line be some definite number expressed to scale in units 
of the load, then a bending-moment scale may be obtained which 
will enable the bending moment to be measured directly from 
the diagram and obviate the necessity of multiplying each 
measurement by the polar distance. The pole, 0, of the force 
diagram of Fig. 84 was taken at a distance of i inch from the 
load line ab, the polar distance oh, therefore, representing 
150 pounds. Then, since the linear scale is i inch = 4 feet, an 
ordinate measuring i inch represents 4 feet; but as this must be 
multiplied by the polar distance we shall have: 

I inch = 4 feet X 150 pounds = 600 pounds-feet, 
or eV inch =10 pounds-feet, 

a new and convenient scale by which the bending moments can 
be measured directly from the diagram. The bending-moment 
scale is derived in each instance by multiplying the linear scale 
by the polar distance expressed in pounds or tons. The ordinate 
yz of the funicular polygon, Fig. 84, measures 21.75 sixtieths of 
an inch, and the bending moment at the section a of the beam is, 
therefore, 21.75 ^ ^o = 217.5 pounds-feet. 

A few examples of the application of the funicular polygon to 
beams will be given. 

Example I. — A beam supported at the ends is 20 feet long, 
weighs 400 pounds, and has concentrated loads of 360 pounds 
and 440 pounds at 8 feet from the left end and 4 feet from the 
right end respectively. Draw the bending-moment and shear 
diagrams, and measure the bending moments under the con- 
centrated loads and the shear stress at the middle of the 
beam. 



FUNICULAR POLYGON 1 27 

Solution. — Select a linear scale of J inch = i foot, and a load 
scale of I inch = 400 pounds. 

Set out the beam as shown in Fig. 85, and since the beam 
weighs 400 pounds it has, in addition to the concentrated loads, 
a uniformly distributed load of 20 pounds per foot. 

We shall first construct the funicular polygon for the concen- 
trated loads, neglecting for the present the uniformly distributed 
load. 

Set off the load line ac by making ah measure 0.9 inch and he 
measure i.i inches to represent the loads of 360 pounds and 
440 pounds respectively. Select the pole at a distance of 1.25 
inches from the load line, so that the polar distance will represent 
1.25 X 400 = 500 pounds. We shall then have for the b ending- 
moment scale, \ inch = i foot X 500 pounds = 500 pounds-feet, 
or iV inch = 100 pounds-feet. 

Draw the vectors, oa, oh, and oc. From a point s in the line of 
action of Ri draw a parallel to oa and produce it until it intersects 
the line of action of Wi at the point r. From r draw a parallel 
to oh, producing it until it intersects the line of action of W2 at 
the point j. From j draw a line parallel to oc and produce it 
until it intersects the line of action of R2 at the point k. Join 
k with s. Then js is th§ closing line of the funicular polygon 
srjk. Draw od parallel to sk. Then cd and da represent in 
magnitude and direction the support reactions R2 and Ri re- 
spectively, and the funicular polygon srjk is the diagram of 
bending moments for the concentrated loads. 

The bending-moment diagram of the evenly distributed load 
of 400 pounds will be parabolic in form, and it will be convenient 
to construct its funicular on the closing line ks of the funicular 
of the concentrated loads. To do so we will take the same pole, 
0, as was used for the concentrated loads, and will consider the 
whole of the distributed load of 400 pounds to be concentrated 
at the middle of the beam, as shown in the figure. This will add 



128 THE ELEMENTS OF MECHANICS OF MATERIALS 



(F) (4oqibs.) (E) 

360 lbs. I 



^ 



440 lbs. 

-4=^—4'- 



0) 



m 




Fig. 85. 



FUNICULAR POLYGON 1 29 

200 pounds each to R2 and Ri. On each side of d in the load 
line lay off ed and df, each one-half inch in length, to represent 
these additions to the support reactions. Draw the vectors 
oe and of. From k and ^ draw parallels to oe and of respectively. 
They intersect at z. Then zsk is the funicular polygon for the 
distributed load, supposing it to be concentrated at the middle 
of the beam. It has been shown in Art. 18 that the bending 
moment at the middle of a simple beam uniformly loaded is only 
one-half that due to the same load concentrated at the middle. 
Hence, the ordinate ti measures the bending moment at the 
middle of the beam due to the distributed load, i being the middle 
point of tz. A parabola constructed on sk as a chord, and passing 
through the point i, is the bending-moment diagram due to the 
distributed load, and srjki is the complete bending-moment 
diagram for the beam. 

The reaction R2 measured on the load line is cd -i- df = H 
inches; hence, R2 = H X 400 = 696 pounds. The reaction 
Ri = da -\- de = H inches; hence, Ri = H X 400 = 504 pounds. 

The ordinates y and y' under Wi and W2 measure ft inch and If 
inch respectively. The bending moments under Wi and W2 are 
therefore 3400 pounds-feet and 2600 pounds-feet respectively. 
These results may be checked easily by calculation. 

Commencing at the left support the shear due to the uniform 

load is equal to half that load, and gradually decreases from left 

to right until, at the middle, it becomes zero, and at the right 

tv L IV L 

support it becomes wL = ? or to one-half the load, 

2 2 

but negative, w denoting the load per unit of length. The dia- 
gram d'e'fd^\ therefore, represents the shear due to the 
distributed load. 

The total shear at the left support is Ri and is equal to da -f de. 
The parallel to e^f shows the gradual decrease of the shear from 
the left support to Wi due to the uniform load. Passing Wi 



130 THE ELEMENTS OF MECHANICS OF MATERIALS 

the shear suddenly drops to n and becomes negative, mn being 
equal to ah. The parallel to e^f drawn from n shows the gradual 
negative increase in the shear from Wi to W2 due to the dis- 
tributed load. Passing W2 the shear suddenly drops to q, pq 
being equal to be. The parallel to e'/' drawn from q shows the 
further increase in the shear due to the distributed load until, 
at V, it becomes — R2 = — (dc -{- df) . The ordinate uw at the 
middle of the beam measures /o inch; the shear at the middle 
section is, therefore, /o X 400 = 56 pounds. 

Example 11. — The beam with overhanging ends of Fig. 52, 
p. 52, loaded uniformly with 20 pounds per foot and with two 
concentrated loads, may be solved readily by means of the 
funicular polygon. The concentrated loads Wi and W2 are 
200 pounds and 400 pounds respectively. 

Set off the beam as shown in Fig. 86. Divide the beam into 
any convenient number of parts, eight in this instance, so that 
each part will be 4 feet in length and will bear 80 pounds of the 
distributed load. These eight parts constitute as many external 
forces, each acting at its center of gravity. 

Letter the beam according to the Bow system, and adopt the 
following scales: Linear, o.i inch = i foot; load, i inch = 400 
pounds. We shall then have: Wi = f§§ = 0.5 inch; W2 = U% 
= I inch; and the load of each of the equal divisions will be 
represented by ^W = 0.2 inch. 

Set off the load line ak accordingly. Select a pole distant 
1.5 inches from the load line, so that the polar distance will 
represent 600 pounds. We shall then have: jV inch = i foot X 
600 = 600 pounds-feet, or ^V inch = 100 pounds-feet, for the 
bending-moment scale. 

Draw the vectors of the force polygon. From some point m 
in the Hne of action of Ri draw a parallel to oa and produce it 
until it intersects the Hne of action of AB at n. Through n 
draw a parallel to oh, producing it to its intersection p with 



FUNICULAR POLYGON 



131 



200 lbs. 



400 lbs. 




Fig. 



132 THE ELEMENTS OF MECHANICS OF MATERIALS 

the line of action of BC, or of Wi. Draw parallels to the other 
vectors as shown. The member qr, parallel to oj^ intersects the 
line of action of JK at r. The parallel to ok through r intersects 
the line of action of R2 at s. The closing member is, then, sm, 
and mnp . . . qrs is the funicular polygon, or the bending- 
moment diagram of the beam. 

Draw ol parallel to sm. Then kl and la represent to scale the 
reactions R2 and Ri respectively. 

It should be noted that ordinates within mnpt and sur give 
negative bending moments, and that t and u are the points of 
inflection. The ordinates y and y give the maximum negative 
and positive bending moments respectively, and as they measure 

-^^ and — of an inch respectively, the bending moments are 
60 60 

1440 pounds-feet and 1600 pounds-feet. 

Commencing at the left end of the beam, the shear increases 
from zero to — ah when Wi is reached. Passing Wi the shear 
becomes — ah — he = — ac, and increases up to the left sup- 
port, where it becomes — {ah -\- he -\- cd) = — ad. Passing the 
left support it becomes Ri — ad = Id and decreases up to 
W2, where it becomes Id — dg = Ig. Passing W2 it becomes 
h ~ g^ = ~ ^^j ^^<i gradually increases until the right support 
is reached, where it becomes — Ih — hz = — Iz (the center of 
gravity of the weight on the seventh division of the beam happen- 
ing to fall directly over the right support, one-half of that weight 
(//) actually lies to the left and one-half to the right of the 
support, and must be so considered) . Passing the right support 
the shear again becomes positive and equal to R2 — Iz = zk, and 
then decreases to zero at the right end of the beam. 

Example III. — To draw the bending-moment and shear dia- 
grams of the cantilever with concentrated loads, as shown in 
Fig. 87, we proceed as follows: 

Draw the load line ad, making ah, he and ed equal, to some 



FUNICULAR POLYGON 



133 



selected scale, to the loads AB, BC, and CD respectively. Select 
some pole 0, and draw the vectors oa, oh, oc, and od. From some 
point r in the support line draw a parallel to oa, producing it 
until it intersects the line of action oi AB Sit q. From q draw a 
parallel to ob and produce it until it intersects the line of action 
of BC at p. From p draw a parallel to oc and produce it until 
it intersects the line of action of CD at n. From n draw a 
parallel to od to meet the line of support at m. Then mnpqr is 




Fig. 87. 



the bending-moment diagram of the cantilever, and the ordi- 
nate under any section of the beam is the measure of the bending 
moment at that section. The maximum bending moment is, 
of course, at the wall. 

The shear diagram is drawn by projection from the load line 
and presents no difficulties. 

In the case of a cantilever it is found convenient to select the 
pole at some chosen perpendicular distance from either extrem- 



134 



THE ELEMENTS OF MECHANICS OF MATERIALS 



ity of the load line. In Fig. 87 the pole might just as well have 
been chosen at 0' and an equal bending-moment diagram, 
mn'p'q'r' , constructed, as shown by the dotted lines. 

Example IV. — The cantilever of Fig. 39, p. 40, having a com- 
bination of concentrated and uniformly distributed loads may be 
solved easily by means of the funicular polygon. 

Using the same linear scale of -^^ inch = i foot as for Fig. 39, 
and a load scale of 400 pounds to the inch, we shall first con- 
struct the funicular polygon for the concentrated loads. 



Reduced to scale, AB = ^W = yf ^ inch, and BC = 4*0*0 = 



JUL 

100 



inch. Set off the load hne ac, Fig. 88, making ab and be 
equal to yf ii^ch and yVV inch respectively. 



(D)(192 lbs.) 




Fh 



Choose a pole at a perpendicular distance of i inch from a, 
so that the polar distance oa represents 400 pounds. We shall 
then have for the bending-moment scale, y^o inch = i foot X 
400 pounds = 400 pounds-feet, whence 5V i^ch = 40 pounds-feet. 
Draw the vectors oa, ob, and oc. 



FUNICULAR POLYGON 135 

Commencing at some point q in the wall line, draw qp parallel 
to oa, pn parallel to oh, and nm parallel to oc, thus forming the 
funicular polygon, or bending-moment diagram, mnpq for the 
concentrated loads. 

The cantilever is uniformly loaded with 48 pounds per foot 
for a distance of 4 feet from the wall, making a total uniform 
load of 192 pounds. Assuming this load concentrated at the 
outer extremity of the 4 feet, set off ad equal to \%% = f| inch 
to represent it (the uniform load AD h taken contraclockwise 
in order to join its funicular to the line qp of the funicular poly- 
gon of the concentrated loads). Draw the vector od, and from 
r, the intersection of pq with the vertical at 4 feet from the wall, 
draw rt parallel to od. It can easily be shown that the bending 
moment due to a concentrated load at the end of a cantilever is 
twice that due to the same load uniformly distributed, so the 
distance qt must be bisected at s, and the parabola having its 
apex at r, and passing through 5, gives rsq as the bending-moment 
diagram due to the distributed load. Then mnprs is the com- 
plete bending-moment diagram of the cantilever. The ordinate 

ms at the wall measures -^^ inch, and the bending moment is, 

50 
therefore, 25.3 X 40 = 1012 pounds-feet, the same as found on 
page 41. 

The shear diagram is drawn by projection from the load line 
and presents no difficulties. 

Example V. — A circular steel axle, supported on end jour- 
nals, is 10 feet long between journal centers and subjected to a 
load of 12,000 pounds at a point 4 feet from the center of the 
journal at the right end. Determine the dimensions of the 
axle. 

Solution. — The axle is to be regarded as being subjected only 
to a bending stress. 

To a scale of f inch = i foot, lay off the axle in skeleton form, 



136 



THE ELEMENTS OF MECHANICS OF MATERIALS 



Fig. 89. To a load scale of | inch = 1200 pounds, lay off the 
load line ah, 1.25 inches in length, to represent the load of 12,000 
pounds. Select a pole one inch from ah, so that the polar 




distance oh represents 9600 pounds. Then, f inch = i foot 
X 9600 pounds = 9600 pounds-feet, or 4V inch = 640 pounds- 
feet, a convenient bending-moment scale. 



FUNICULAR POLYGON 137 

Draw the vectors oa and oh. Commencing at a point m in 
the Hne of action oi Ri, draw the funicular polygon mjk. Draw 
oc parallel to the closing Hne km. Then he and ca are the re- 
actions R2 and Ri respectively, in magnitude and direction. By 
scale measurement he = 7200 pounds = R^^ and ca measures 
4800 pounds = Ri. 

Treating the journals as uniformly loaded cantilevers, we have 

R 7 

-^- as the maximum bending moment of the journal at the left 

end, in which I denotes the length of the journal. 

The resisting moment is — = - — = = 0.196 Sd^, in which 

c 04 c 32 

d denotes the diameter of the journal. From this it is seen that 
the resisting moment, and therefore the bending moment, is pro- 
portional to the cube of the diameter. 

Then, — ^ = o.iq6 Sd^, whence d = y ^ • - • 

2 ^ 0.1965 d 

It is usual to fix the ratio — in accordance with the conditions 

d 

of the case. Assuming the direction of the load to be constant 

and the maximum revolutions to be 250 per minute, it is good 

practice to make — = 2 ; for slower speeds the ratio would be less. 
d 

Then, taking 5 at 10,000 pounds per square inch, we have 



d = 0.0226 V4800 X 2 = 2.21 inches, say 2 J inches; 
whence ^ = 42 inches. 

For the journal at the right end, 



d' = 0.0226 V7200 X 2 = 2.71 inches, say 2| inches; and 
r = 5.5 inches. 

Since the bending moment at any section of the axle is pro- 
portional to the cube of the diameter at the section, it follows 
that the diameter at any section is proportional to the cube root 



138 THE ELEMENTS OF MECHANICS OF MATERIALS 

of the bending moment at the section. The lengths / and V 
being known, the bending moments M and M' may be measured 

7? 7 7? 7' 

from the diagram or calculated from — - and — ^ respectively. 

2 2 

These moments are found to be: M = 625 pounds-feet, and 
M' = 1200 pounds-feet. The bending moment Mi at the load 
measures 28,800 pounds-feet. 



Then, |^ = ^ = ^§ = ^^ 

d^ 2.71 ^ M' ^ 1 



28,800 

- =2.< 



2.71 ^ ivi ' 1200 

whence di = 7.8 inches, say 7x1 inches. 

Making the hub seat of the wheel 8 inches long, we have now 
to determine ^ and d^. The bending moments M2 and if 3 
measure 26,240 pounds-feet and 27,200 pounds-feet respectively. 



Then, I = ^ = ^?6.M2 ^ ..^^g, 

a 2.71 ' 1200 

whence ^ = 7.577 inches, say 7! inches. 

1 ds ds .727,200 

and -=• =\/-^-- = 3.517, 

^ 2.21 '^ 625 

whence J3 = 7.773 inches, say 7! inches. 



An empirical rule in machine design makes the height of the 

collar of the journal f- -inch, and its width 1.5 times the height. 

10 8 

Hence, 

221 I II . 
Height of journal collar at left end = -^ — + - = — inches, 

10 8 32 

whence width = Jl inches. 

27^ I "i 
Height of journal collar at right end = -^-^ + o "= o inches, 

10 o 8 

whence width = f^ inch. 

The axle is drawn in Fig. 90 to the scale of one-sixth. 
Example VI. — The total weight on the 4 axles of a semicon- 
vertible street car is 46,560 pounds. Motion is conveyed to 



FUNICULAR POLYGON 139 

the axles from the motors, one for each axle, by means of 
gears, and the weight is carried on journals that overhang the 
car- wheel journals 9.25 inches. The horse power of each motor 
is 40, and the maximum speed on level is 25 miles per hour. The 
diameter of the car wheels is 33 inches and the gauge of the road 
5 feet. Determine the dimensions of the axles. 

Solution. — A car axle is subjected to bending only when the 
car is moving on a straight portion of the road, but in rounding 
curves it is subjected to both torsion and bending, so that the 
equivalent twisting moment must form the basis for the deter- 
mination of the dimensions of the axle. 

The load carried on each journal is — - = ^—~ — = 5820 pounds, 

which occasions an equal reaction at the journal of each car 
wheel. The condition then is one of a beam having equal over- 

W 
hanging ends, each overhang being uniformly loaded with — or, 

8 

W 
which is the same thing, having — concentrated at its middle. 

8 

To a linear scale of f inch = i foot, or of one- twentieth, lay off 
the axle in skeleton form. Fig. 91. To a load scale of i inch = 582 
pounds lay off ah 1.25 inches in length to represent the load AB 
of 5820 pounds, and he the -same length to represent the equal 
load BC. Select a pole 0, distant 1.25 inches from the load 
line ac, so that the polar distance will represent 5820 pounds. 
Then, f inch = i foot X 5820 pounds = 5820 pounds-feet, or 5V 
inch = 194 pounds-feet, which is a convenient b ending-moment 
scale. 

Draw the vectors oa, oh, and oc. Commencing at some point 
m in the line of action of i?i, draw ma parallel to oa, and from a, 
the point of intersection of ma with the line of action oi AB^ 
draw an parallel to oh, and from n, its intersection with the line 
of action of BC, draw nk parallel to oc, intersecting the hne of 
action of Ri at k. Join k with m. Then mank is the funicular 



I40 THE ELEMENTS OF MECHANICS OF MATERIALS 




Fig. 92. 



FUNICULAR POLYGON 141 

polygon, or b ending-moment diagram., of the axle. Draw od 
parallel to km\ it will, of course, coincide with oh, since ah and 
he are equal and was taken on the horizontal through h. 
The bending moment M between the supports is constant and 

measures -^^ X 194 = 4500 pounds-feet. 
50 
To find the twisting moment we must know the revolutions 
per minute of the axle. 

Circumference of car wheel = — ^ = 8.64 feet. 

12 

o J r • . 5280 X 25 , ^ 

Speed of car per mmute = - — ~ == 2200 feet. 

2200 ^ 

Pr^M,= 40X33,000 ^ pounds-feet. 
2 X 3-1416 X 254.63 ^ ^ 

To find the equivalent twisting moment (see Art. 47) we have 

Me=M-\- VW^^M} = 4500 + V(45oo)2 + (825)2 
= 9075 pounds-feet = 108,900 pounds-inches. 

Taking S as 6000 (see Art. 46), we shall have 

108,900 = o.ig6^Sd^ = 0.196 X 6000 d^, 
in which d is the diameter of the axle. 



Then d = y '-^ = 4.52 inches, say 44 inches. 

Treating the overhanging journals as cantilevers uniformly 

loaded, or with the load — concentrated at the middle, the 

8 

maximum bending moment is M = 4500 pounds-feet. For the 
diameter of the journals we shall then have 



4500 X 12 = 0.1965^1^ whence di= \/^^^^~ =3-5S inches 



142 THE ELEMENTS OF MECHANICS OF MATERIALS 

Or, since the bending moments are proportional to the cubes 
of the diameters, we shall have, denoting the diameter of the 
journals by di, 



-~- = U^—^ = 1.26, whence di = 3.59 inches, say 3! inches. 



4500 

Denoting the length of the journals by /, and taking the ratio 

—-as 2, we shall have 
di 

Length of journal = 3.59 X 2 = 7.18 inches, say 7A inches. 

The height of collar at outer end of journal = ^^ + - = 

10 8 

0.484 inch, say J inch. 

Width of collar = 0.484 X 1.5 = 0.7 inch, say H inch. 

The diameter of the axle has been determined under the 
supposition that the gear wheel which receives the power from 
the motor is fitted to the axle by hydraulic pressure. Should 
the gear wheel be keyed to the axle the diameter would have to 
be increased by an amount equal to the depth of the key way, 
amounting in this instance to about t^ inch. 

The axle is drawn in Fig. 92 to the scale of one-sixth. 

Example VII. — A steel axle rests on two journals and is 
subjected to a vertical load of 6000 pounds applied on an over- 
hanging end at a distance of 15 inches from the center of the 
nearest journal. The distance between the journals is 4 feet. 
Construct the funicular polygon, and determine the magnitude 
and direction of the reactions. 

Solution. — To a linear scale of f inch = i foot lay off the 
axle in skeleton form. Fig. 93. To a load scale of i inch = 
1500 pounds lay off the load line ab one inch in length to repre- 
sent the load AB oi 6000 pounds. From the pole 0, distant i 
inch from ab, draw the vectors oa and ob. The polar distance 
of I inch represents 6000 pounds to scale, and we shall have 

finch = I ft. X 6000 lbs. = 6ooolbs.-ft., or 4V inch = 2oolbs.-ft. 

as a bending-moment scale. 



FUNICULAR POLYGON 



143 



From a point m in the line of action of Ri draw a parallel to 
oa, intersecting the line of action of W at j. From j draw a 
parallel to oh^ intersecting the line of action of R2 at k, and from 
k draw km as the closing line of the funicular polygon mjk. 
From draw a parallel to km, intersecting the load line pro- 
duced at c. Then be and ca represent the reactions R2 and Ri 
in magnitude and direction respectively. By scale measurement 
be = R2 = 7875 pounds, and ca = Ri = 1875 pounds. 




Fig. 93- 



PROBLEMS 

1. A beam 22 feet long supports a load of 1000 pounds at a point 6 
feet from the left end and one of 1200 pounds at 5 feet from the right end. 
In addition there is a uniformly distributed load of 100 pounds per foot. 
Construct the bending-moment and shear diagrams. What concentrated 
load must be placed at the middle of a similar beam in order that the 
maximum bending moment shall be the same as that of the given beam ? 

Ans. 2190 lbs. 

2. A beam 30 feet long weighs 20 pounds per foot and overhangs each 
support 6 feet. It bears a superimposed load of 100 pounds per foot, and 
a load of 1400 pounds concentrated at a point 3 feet to the right of the 



144 THE ELEMENTS OF MECHANICS OF MATERIALS 

middle. Construct the bending-moment and shear diagrams, and find 
graphically the bending moment at the dangerous section and the dis- 
tances of the points of inflection from the left support. 

Ans. 7760 lbs. -ft.; 1.48 ft. and 16.9 ft. 

3. A cantilever 14 feet long supports three concentrated loads,' — 500 
pounds at 4 feet from the wall, 600 pounds at 10 feet from the wall, and 
200 pounds at the extremity. In addition it bears a uniformly distrib- 
uted load of 80 pounds per foot run. Construct the bending-moment and 
shear diagrams. 

4. Determine the diameters of the journals of the axle of Example VII, 
page 142. Ans. if ins. and 4^ ins. 



CHAPTER VIII 
FRAMED STRUCTURES. RECIPROCAL DIAGRAM 

60. Framed Structures. — A framed structure is an assem- 
blage of members for the transmission or modification of external 
forces, the internal stresses occasioned thereby in the members 
being principally those of tension and compression. A member 
in tension is known as a tie; if in compression, it is known as a 
strut. 

A frame is a theoretical structure, the joints connecting its 
members being supposed frictionless. There are, of course, no 
such things as frames, since all joints offer some resistance to 
rotation. In general, however, engineering structures approach 
so nearly to frames that no sensible error results from treating 
them as such. 

The members of a frame are rigid bars, hinged at the ends, and 
it is assumed that : {a) The pins at the joints are without friction. 
(5) The external forces acting on the frame are appHed at the 
joints. 

If the point of application of a load be at some point inter- 
mediate between the joints, parallel forces equivalent to the load 
must be substituted at the joints. If the point of appHcation 
be midway between joints, the equivalent parallel forces at the 
joints will each be one-half the load; if the point of application 
be otherwise than at the middle, then the equivalent parallel 
forces at the joints may be found by moments. 

61. Loads. — The load on a structure consists of the weight 
of the structure itself and the external forces acting on it. 
Stationary and moving weights constitute dead and live loads 

145 



146 THE ELEMENTS OF MECHANICS OF MATERIALS 

respectively. Wind pressure, the weight of the covering of a 
structure and the weights the structure may support, are external 
forces. The reactions of the supporting foundations of a struc- 
ture are external forces, but they are distinguished from the 
external forces constituting the load by calling them supporting 
forces. 

For the equilibrium of a structure the external forces constitut- 
ing the load must equal the supporting forces, and there must 
be a balance between the external and the internal forces. 

62. Trusses. — Trusses are frames designed to support the 
roofs of buildings and such loads as are carried by bridges, the 
supports being widely separated. There are two classes of 
trusses: Those in which the upper and lower members, called 
chord members, are parallel and horizontal; and those whose 
chords are not parallel. The members connecting the upper 
and lower chords are known as braces, or as web members, and 
may be vertical or diagonal. The points at which web members 
meet a chord divide the truss into bays or panels, and the measure- 
ment of a bay is the horizontal distance between its joints. 

The triangle being the polygon whose shape cannot be changed 
without altering the length of its sides, all bridge and roof trusses 
are made up of triangular frames as the best means of securing 
rigidity. 

Roof trusses are placed from 10 to 16 feet apart, and each 
truss is known as a principal. The upper chord is sometimes 
known as the principal rafter. 

63. Distinction between Beams and Girders. — Framed 
structures, such as are used in bridges and for the support of 
roofs, are beams in the sense that they are supported at the ends 
and carry their loads between the supports, but the term beam 
seems to be restricted to the cases where its application is of 
the simple form and of solid section. Thus, the beam of I sec- 
tion, when used alone, is known as a beam, but when two of them 



FRAMED STRUCTURES — RECIPROCAL DIAGRAM 147 

are compounded with plates riveted to their top and bottom 
flanges, the combination is known as a box girder, the section 
of which is not soHd. Generally speaking, beams of built-up 
section are girders. 

A truss with its upper and lower chords parallel and hori- 
zontal is a direct transformation from the simple I beam, the 
chord members, like the flanges of the /, resisting the bending 
moment; the web members, like the web of the /, resisting the 
vertical shear and transmitting it from member to member to 
the supports. 

If a chord of a truss is inclined to the horizontal it aids in the 
transmission of the vertical shear and is not, therefore, designed 
only to resist the bending moment. 

64. Force Action at a Framed Joint. — A structure can remain 
in a state of rest only when there is equilibrium in the system of 
forces acting on it, and in order that there shall be a state of 
rest at a joint there must be equilibrium in the forces acting 
on it or transmitted to it. Since the joints are to be considered 
frictionless, the external force acts through the center of the joint, 
and the action of a member on the pin is balanced by the reac- 
tion of the pin on the member, , ^ 

the action and reaction being \ / 

' \ / 

normal to the surface of con- \ / 

tact; and since the joints are in \ / 

equilibrium, the stresses in the \^/ ^ ^ 

ends of a member must be equal /\ 

and opposite, therefore the lines / \ 

of action of the stresses in the 7 \ 

members lie in the straight lines / \ 

joining the centers of the pins. 

If P be an external force act- ^^' 

ing at the joint A, Fig. 94, its components along the members 
AB and AC (shown by the dotted lines) are equal and opposite 



148 THE ELEMENTS OF MECHANICS OF MATERIALS 

to the stresses in the members. In all cases the external force 
at a joint of a frame must be in equilibrium with the stresses 
in all the members which meet at the joint. 

65. Reciprocal or Stress Diagram. — Since there is equilib- 
rium at each joint of a framed structure, it follows that a 
closed polygon may be constructed whose sides represent the 
acting forces in magnitude and direction. Such a polygon is 
known as the stress diagram of the frame, but, owing to its inter- 
changeable relations with the frame, it is also known as the 
reciprocal diagram of the frame, and is nothing more nor less 
than a polygon of forces. 

66. Frame Diagram. — As a preliminary to the construction 
of the reciprocal diagram a scale drawing of the framework must 
be made, which is known as the frame diagram. The complete 
preparation of the frame diagram comprises the following: 

1. The determination of the magnitude and direction of the 
total load at each joint, replacing any load that may be applied 
between two joints by equivalent parallel forces at the joints. 
These equivalent parallel forces are determined in the same 
manner as that employed in determining the support reactions 
of a loaded simple beam. 

2. The determination of the supporting forces, or support 
reactions. The reactions can be, and often are, determined 
from the funicular polygon, but their predetermination affords 
a check as to the accuracy, and not infrequently furnishes the 
known external force at a joint where but two members meet, 
thus providing a starting point for the construction of the 
reciprocal diagram. 

3. The correct lettering of the diagram according to the Bow 
system. 

4. The marking with arrowheads of all the external forces, 
giving to each its value, and taking care that the arrowheads do 
not cross the lines of the frame diagram. 



FRAMED STRUCTURES — RECIPROCAL DIAGRAM 



149 



67. Support Reactions. — With a small span the ends of a 
roof truss are fixed, and the reactions at the supports are vertical 
when the loads are vertical. In case of a wind load, acting only 
on one side of the roof, the reactions due to it act in directions 
parallel to the normal wind pressure. 

With large spans one end of the truss is fixed while the other 
end is on rollers, thus permitting a lateral movement in case of 
expansion. The reaction at the fixed end will be inclined and 
that at the free end will be vertical. 

It should be noted that when all the external forces are ver- 
tical the support reactions are vertical. In such cases the loads 
that come directly over the supports are omitted, as they have 
no influence on the stresses in the members. Such loads must 
be deducted from the total support reactions in order to obtain 
the proper reactions to use in the determination of the stresses. 





1+2=3 


tons 








> 


^ 






1+1=2 tons B/^ 

1 y^ 


X 


2+2=4 tons 




It 


on ji^ ^ 


H 


2t 


ons 






^ 


1 \ 


\' 






E 










Ri=4 tons 








R2=5 tons 



Fig. 95. 



For example. Fig. 95 represents a king-post truss with a uni- 
form load of 8 tons on one side and 4 tons on the other. The 
loads are applied to the joints as follows: 

Of the 4 tons on the left-hand rafter we may assume one-half 
of it borne by the member AF and the other half by the member 
BG. Distributing these two loads, giving half of each to the 
ends of the member which supports it, we get a total load of 



I50 



THE ELEMENTS OF MECHANICS OF MATERIALS 



2 tons at joint i, i ton at the apex end of the rafter, and i 
ton over the left support. Proceeding in a similar manner with 
the right-hand rafter, we get a total of 3 tons at the apex of the 
rafters, 4 tons at the joint 3, and 2 tons over the right support. 
To find the support reactions we reject the loads over the 
supports, and, assuming the span to be a, take moments thus: 



i?iXa = 2X^ + 3X- + 4X- 

4 2 



whence Ri = 4 tons. 



i?2Xa=4X^ + 3X- + 2X 



whence R2 = S tons. 



4 

+ 2X^ 

42 4 

These values of Ri and R2 are to be used in determining the 
sti:esses in the members, but the total wall reactions are 5 tons 
and 7 tons for Ri and R2 respectively. 

68. Wind Pressure. — For purposes of computation the direc- 
tion of the wind is assumed to be horizontal, and its intensity 
may be taken as 40 pounds per square foot. 

Roofs generally present an inclined surface to the wind, but 
as the roof structure itself must resist the normal pressure to 
which it is subjected, it is important to know this normal pres- 
sure for the different degrees of roof inclination. 

The normal pressures due to a horizontal intensity of 40 pounds 
per square foot on a vertical surface have been determined experi- 
mentally for roofs of different inclinations, and are set forth in 
the table which follows: 



Pitch of roof in 


Normal pressure in pounds 


degrees. 


per square foot. 


5 


5 


10 


10 


15 


14 


20 


18 


25 


22 


30 


26 


35 


30 


40 


33 


45 


36 


50 


38 


55 


39 


60 


40 



FRAMED STRUCTURES — RECIPROCAL DIAGRAM 151 

For horizontal pressures other than 40 pounds the normal 
pressures are directly proportional to those given in the 
table. 

69. Drawing the Reciprocal Diagram. — In drawing the recip- 
rocal diagram we proceed as follows: 

1. Select a scale — as large as practicable — and draw the 
force diagram of the external forces, marking the beginning and 
ending of a line representing a force with the small letters of the 
alphabet corresponding to the capital letters, taken in clockwise 
order, found in the spaces flanking the force in the frame diagram. 
This diagram represents the magnitudes and directions of the 
external loads on the joints and of the support reactions, and, 
if properly drawn, forms a closed polygon. In the most frequent 
case, that of vertical loads, the polygon resolves itself into a 
straight hne, vertical in direction, as already explained. Deter- 
mine the magnitude of the support reactions, either by moments 
or by the method of the funicular polygon. 

2. Choose as a starting point a joint where a sufficient number 
of conditions are known to enable its reciprocal to be drawn, 
being careful to letter the lines of the reciprocal with the small 
letters of the alphabet corresponding to the capital letters 
denoting the members iru the frame diagram. The lengths of 
the lines of this reciprocal, to the chosen scale, give the magni- 
tudes of the stresses in the members to which they refer, and the 
directions, or the kinds, of these stresses are at once determined 
by noting the directions in which the successive lines of the 
reciprocal were drawn. The directions of the stresses thus 
found are at once indicated by placing arrowheads on the mem- 
bers in the frame diagram. Since there are equal and opposite 
stresses in the ends of a member, arrowheads must now be placed 
on the other ends of the members whose stresses have been 
found, making them point in the opposite direction to those 
placed at the joint whose reciprocal has been drawn. 



152 



THE ELEMENTS OF MECHANICS OF MATERIALS 



3. Proceed by similar processes until the reciprocals of all 
the joints are drawn and the stresses in all the members 
determined. 

It should be noted that the reciprocal of a joint cannot be 
drawn if the stresses in more than two of the members forming 
it are unknown. In general terms, the drawing of the reciprocal 
of a joint of n forces involves 2 n conditions, viz., n magnitudes 
and n directions, and unless 2 n — 2 of these conditions are 
known the data is insufhcient to draw the reciprocal. 




3 tons 



Rip=3 tons 



Fig. 96. 



Example I. — Suppose each rafter of the roof truss of Fig. 96 
to support an evenly distributed load of 6 tons; it is required 
to find the magnitudes and kinds of stresses in the members of 
the truss. 

We commence by apportioning the loads to the joints of the 
frame diagram. In doing so, we find a total load of 6 tons at 
the ridge and a load of 3 tons directly over each supporting wall. 
As these latter do not affect the stresses in the members they 
will be omitted from any further discussion, remembering that 
if the total reactions at the walls are required, Ri and R2 must 
each be increased by 3 tons. 



FRAMED STRUCTURES — RICIPROCAL DIAGRAM 153 

The truss being symmetrical, the support reactions Ri and R2 
are each equal to half of the load of 6 tons at the ridge. 

Letter the frame diagram as shown, ignoring the vertical loads 
over the supports, so that R2 will be known as BC and Ri as CA . 
Adopt a load scale of 0.25 inch to the ton. 

Since each of the three joints of the frame has but two members 
and a known external force, either may be used as a starting 
point. Selecting the joint at the left support, we draw ca equal 
in length to 0.75 inch to represent the left reaction CA of 3 tons, 
and we measure it upward from c to a because Ri acts upward. 
From a and c draw lines parallel to the members AD and DC 
respectively. They intersect at d, giving the triangle cad as 
the triangle of forces, or the reciprocal diagram, for the joint at 
the left support. Hence, ad and dc represent in magnitude and 
direction the stresses in the members AD and DC. Knowing 
the direction of Ri, as represented by ca, the directions of the 
actions of the stresses in the two members are known by taking 
the sides of the triangle in order, as shown by the arrows. The 
stress in AD acts in the direction ad, and that in DC in the 
direction dc. Indicate these directions by placing arrowheads 
on the two members at points close to the joint. Knowing that 
the stress in one end of ajnember is opposed by an equal and 
opposite stress in the other end, we may now place arrowheads 
at the other ends of the members AD and DC accordingly, as 
shown. 

To draw the reciprocal of the joint at the right support we 
draw be upward, and make it 0.75 inch in length to represent 
the 3 tons of the right reaction BC in magnitude and direction. 
From c and b draw lines parallel to the members CD and DB 
respectively. They intersect at d, giving the triangle bed as 
the reciprocal of the joint at the right support. Hence, cd and 
db represent in magnitude and direction the stresses in the mem- 
bers CD and DB. The arrows within the triangle bed show the 



154 



THE ELEMENTS OF MECHANICS OF MATERIALS 



direction of these stresses, and they are marked with arrowheads 
accordingly in the frame diagram. By scale measurements of 
the reciprocals the stresses in the rafters are found to be 4.4 tons 
each, and in the tie CD the stress is found to be 3.2 tons. 

If to the frame of Fig. 96 we add a vertical rod, called the king- 
post, we obtain the frame of Fig. 97. 




Ri= 3 tons 



Ro=3 tons 



Fig. 97. 



To draw the reciprocal of the joint ABED at the ridge we 
measure ah downward, and make it 1.5 inches long to represent, 
to the chosen scale of 0.25 inch to the ton, the magnitude and 
direction of the external force of 6 tons. The stress in the 
rafter AD has been found to be 4.4 tons, so we draw ad parallel 
to AD and make it i.i inches long to represent the 4.4 tons to 
scale. From h we draw a line parallel to BE and we find it 
intersects ad at d. This shows that d and e are one and the 
same point, and that there is no stress in the vertical post DE. 
The purpose of DE is to prevent sagging in the horizontal 
member. 

As a further illustration we will consider the king-post truss 
of Fig. 95, reproduced in Fig. 98. The loads and support reac- 
tions were found to be as shown. 

Knowing the force EA, or Ri, we commence at the joint at 



FRAMED STRUCTURES — RECIPROCAL DIAGRAM 155 




STRESSES 
EF = 6,9 
HG= 3.0 V Ties 
IE = 8.7J 
AF = 8.0' 
BG^e.O 

GF=2.0> struts 
CH = 6.0 
Dl =10.0 
IH=4,0. 



Fig. 99 



156 THE ELEMENTS OF MECHANICS OF MATERIALS 

the left support and, with a scale 0.25 inch to the ton, obtain eaf 
as the triangle of forces for the joint. Marking with arrowheads 
the kinds of stresses in the members AF and FE, we proceed to 
the next joint in clockwise order, that at which the load of 2 tons 
is applied. 

Of the eight conditions of this joint the four directions and 
two of the magnitudes are known, the magnitude of FA having 
just been determined. We then readily obtain ahgf as the force 
polygon of the joint. Marking with arrowheads the kinds of 
stresses in the members BG and GF, we proceed in clockwise 
order to the other joints, obtaining for the joint at the ridge the 
force polygon gbch\ for the joint ECDI the polygon hcdi\ and 
for the joint at the right support the triangle ide, thus completing 
the determination of the stresses in all the members. 

It will be observed that each of the force polygons of the 
joints, when taken in order, contains a side of the one immediately 
preceding it; hence, each polygon can be built upon the one pre- 
ceding it, and so produce one figure which will contain all the 
sides and be a graphic representation of the magnitudes and direc- 
tions of all the external forces and internal stresses of the struc- 
ture. Such a figure is the reciprocal diagram of the structure.^ 

The reciprocal diagram of the truss of Fig. 98 is shown in 
Fig. 99, the load line ad having first been drawn and the lines 
representing the stresses then drawn in their regular order. 
The magnitudes of the stresses were measured to the scale and 
found to be as shown in the table. 

Any attempt to put arrowheads on the reciprocal diagram to 
indicate the kinds of stresses in the members results in nothing 
but confusion. If the kind of stress in a member cannot be dis- 
covered by the eye, the polygon of forces for the joint in question 
must be drawn, as was done in connection with Fig. 98. 

70. Rule for Determining the Kind of Stress in a Member. 
— If, after determining the directions of the stresses in all the 



FRAMED STRUCTURES — RECIPROCAL DIAGRAM 1 57 

members by means of the reciprocals of the joints, and after 
marking the ends of the members with arrowheads accordingly, 
it is found that the arrowheads of a member point toward its 
joints, the member is then in compression and is a strut ; if they 
point away from the joint the member is in tension and is a tie. 
Thus it is found that the rafters AD and BD, Fig. 96, are in 
compression, and the member CD is in tension. 

71. Method of Sections in Determining Stresses. — This 
method depends upon the principle demonstrated in Art. 58, 
that at any imaginary section of a frame there is equilibrium 
between the external forces on one side of the section and the 
stress forces in members on the same side that are cut by the 
section, and that, therefore, the algebraic sum of the moments 
about any point in the plane of the frame must be zero. There 
must, of course, be but one unknown force in the equation of 
moments, and this will be the case: 

(a) When only two members are cut and one of them passes 
through the center of moments. 

(b) When three members are cut and the center of moments is 
taken at the intersection of two of them. 

(c) When the stresses are known in all the members cut 
except one. 

For example, the roof truss of Fig. 98 is reproduced in Fig. 100. 
Knowing the magnitude and direction of the left reaction i?i, 
and considering the equilibrium of the joint at the left support, 
it is seen from inspection that AF is in compression and FE 
in tension, and they are so marked with arrowheads in the 
frame. 

There being equiHbrium at the joint, we have the static equa- 
tions, 6*1 cos 30° = S2 and 6*1 sin 30° = Ri, in which Si and 6*2 
are the stresses in AF and FE respectively. From these equa- 
tions we get 

Si = stress in AF = 8 tons, and ^2 = stress in FE = 6.9 tons. 



158 THE ELEMENTS OF MECHANICS OF MATERIALS 

Knowing the stress in AF and the external force AB, an inspec- 
tion of joint ABGF shows BG and OF to be struts, and they are 
so marked in the frame. The section xy cuts the members BG, 
GF, and FE, two of these members intersecting at E. Consider- 
ing the part of the truss to the left of xy, and denoting the stress 
in BG by 6*3, we have, by moments about the joint at the middle 
of the lower chord, 

53 X 12.5 sin 30° + 2 X 6.25 = 4 X 12.5, whence Sz = 6 tons. 



3 tons 



4 tons 




Sj cos 30' 



R2=5 tons 



Fig. 100. 



Calling Si the stress in GF, and taking moments about the left 
support, we have 

54 X 12.5 sin 30° = 2 X 6.25, whence 6*4 = 2 tons. 

These results are the same as were obtained from the reciprocal 
diagram of Fig. 99. 

In the examples of the section method just given the nature 
of the stress, whether tension or compression, in the member 
whose stress was sought was known, so that the nature of the 
moments, whether clockwise or con traclock wise, about the 
center of moments was known. It is not, however, essential 
that the nature of the stress in a member be known in order to 



FRAMED STRUCTURES — RECIPROCAL DIAGRAM 159 

determine it. A member may be assumed to be either in ten- 
sion or in compression and the equation of moments written 
accordingly. Should the solution of the equation give a positive 
result for the stress, the assumption of its nature is the correct 
one; should the resulting stress be negative in sign, then the 
assumption as to the nature of the stress is incorrect, but the 
numerical value of the stress will be correct and the same as in 
the first instance. These conditions arise from the fact that, 
in the first instance, the moment of the required stress was 
placed in the proper member of the equation of moments; in 
the second instance it was improperly placed. 

For example, in finding the stress in BG we will assume it to 
be one of tension. The equation of moments about the middle 
of the lower chord as a center will be 

4 X 12.5 + 5*3 X 12.5 sin 30°= 2 X 6.25, whence ^3= — 6 tons, 
a negative result, which shows that BG is in compression and not 
in tension, as was assumed, but the numerical result is the same 
as found above. 

Again, assuming FE to be in tension, we shall have, with 
moments about the joint at the middle of the left rafter, 

6*2 X 6.25 tan 30° = 4 X 6.25, whence ^2 = 6.9 tons, 

as was found above, the positive result showing that FE was 
correctly assumed to be in tension. 

72. Stresses in Braced Cantilevers. — The stresses in the 
members of the braced cantilever of Fig. loi, having a con- 
centrated load of 2 tons at its outer extremity, may be found 
as follows: 

To the scale of 4V inch = iV ton lay off ah vertical and i inch 
in length to represent the load of 2 tons. From h draw a parallel 
to BC and from a a parallel to CA . They intersect at c. Then, 
abc is the triangle of forces for the equilibrium of the joint A BC. 
The force AB, acting downward, is denoted in direction and 



l6o THE ELEMENTS OF MECHANICS OF MATERIALS 

magnitude by ah in the triangle, and the stresses in the other 
members taken in order act in the directions he and ca, and their 
magnitudes are, of course, denoted by the lengths of these lines 
to the chosen scale. 




2 tons 



d ' 7i 
Fig. loi. 



Mark these directions by arrowheads at the joint ABC. The 
directions of the stress in the upper end of BC and in the outer 
end of CA being now known, it follows that the stresses in the 
other ends of these members are equal and opposite, and arrow- 
heads must be at once placed to indicate them. We now have 
sufficient data to draw the force diagram of the joint CBD. The 
stress in the lower end of CB has just been marked to act in the 
direction cb. From b and c draw parallels to BD and DC re- 
spectively, intersecting at d; then the triangle cbd is the force 
diagram for the joint CBD, and bd and dc are the directions of 
the stresses in the members BD and DC, and they have been 
marked accordingly in the frame diagram. Arrowheads are at 
once placed at the other ends of BD and DC to denote the 
directions of the equal and opposite stresses at the wall and at 



FRAMED STRUCTURES — RECIPROCAL DIAGRAM i6l 

the joint ACDE respectively. The stresses m AC and CD at 
the joint ACDE are now known and their directions are ac and 
cd respectively. From d draw a parallel to DE and from a 
a parallel to EA. They intersect at e, and de and ea are the 
directions and magnitudes of the stresses in the members DE 
and EA respectively. It should be noted that, in the reciprocal 
diagram, the stress line of one member may lie wholly or partly 
on the stress line of another member, as was here instanced in 
the stresses of ^C and EA. The stresses in the different mem- 
bers were measured to scale on the reciprocal diagram and marked 
on the members of the frame. 

The horizontal outward pull of 2.95 tons at the upper joint at 
the wall occasions an equal and opposite reaction. At the lower 
wall joint there is a horizontal thrust of 2 tons and a diagonal 
thrust of 2.25 tons. The horizontal component of this diagonal 
thrust is hd = ac = 0.95 ton, making the two reactions equal 
but opposite in direction. 

The braced cantilever of Fig. 102 is 25 feet long, 10 feet deep 
and uniformly loaded on the top with 100 pounds per foot run. 

In apportioning the total load of 2500 pounds, it will be 
observed that the member BJ, being but half the length of each 
of the members CH and DF, sustains but one-fifth of the total 
load, CH and DF each sustaining two-fifths. 

The apportionment of the load will therefore be: 500 pounds 
at the joint at the outer end, 1000 pounds at the joint CDFGH, 
750 pounds at the joint BCHIJ, and 250 pounds at the joint 
ABJ at the wall. This latter load has no influence on the 
stresses in the members and is rejected. 

Commencing at the joint DEF, and with a scale of 2V inch 
= 100 pounds, we get def as the force diagram for the equilibrium 
of the joint, and at once mark arrowheads on the frame indicating 
the directions of the stresses EF, FD, FE, and DF. Having now 
the magnitude of the stress FE at the joint FEG, we obtain 



l62 



THE ELEMENTS OF MECHANICS OF MATERIALS 



feg as the force diagram of the joint. Marking with arrowheads 
on the frame the directions of the stresses thus obtained, we find 
that at the joint CDFGH the force CD and the stresses in DF 
and FG are known, and therefore the force polygon cdfgh is 
readily obtained. The reciprocal diagram may now be com- 
pleted without difficulty. 



1000 lbs 



600 lbs 




Fisr. I02. 



73. The Warren girder of Fig. 103 has a span of 27 feet, and 
is loaded with ij tons per foot, making a total load of 36 tons. 
The members AF and DL each sustains but one-sixth of the 
load, and, in the apportionment, one-twelfth, or 3 tons, falls 
over the support at each end, and the remainder of the load as 
shown. 

At each end of the top flange there is equilibrium under the 
action of the external force of 3 tons in line with the upright 
member and the stresses in the two members at right angles to 
each other. Evidently the stress in each of the uprights is 



FRAMED STRUCTURES — RECIPROCAL DIAGRAM 



163 



3 tons, therefore there cannot be any stress in the members AF 
and DL if the equihbrium is to be maintained. To indicate 
that there is no stress in AF the letters a and / must be placed 
at the same point in the reciprocal diagram, and because of the 
absence of stress in DL the letters d and I are at the same point. 
It should not be forgotten that the two external forces of 3 tons 
at the ends, coming directly over the supports, are neglected, and 



3]tons 9 jtons 



tons 3 .tons 



15 tons 




--p-S sin 



STRESSES 
FG = KL=17.32tons 
GE=KE = 8.66 tons 
GH = H I = I J ^J K=7.5 tons 
E 1^16.2 tons 
BH=CJ=12.5toias 



Fig. 103. 



that the reactions of 15 tons each are due solely to the loads of 
9 tons, 12 tons, and 9 tons, the forces which occasion the stresses 
in the members. In the construction of the Warren girder the 
end uprights and the members AF and DL are omitted. 

To the scale of ^ inch = 3 tons set off ef to represent in mag- 
nitude and direction the left reaction EF of 15 tons. From/ 
and e draw parallels to FG and GE respectively, intersecting at g. 
Then, efg is the triangle of forces for the joint at the left support. 



164 



THE ELEMENTS OF MECHANICS OF MATERIALS 



For the joint GFABH we get gfabh as the polygon of forces, and 
so on to the completion of the reciprocal diagram. 

The triangles of the Warren girder being equilateral, the stresses 
found from the reciprocal diagram may easily be checked. Thus, 
denoting the stress in LK by S, we have 

5 = 15 sec 30° = 17.32 tons. 

74. The Linville or N girder of Fig. 104 is irregularly loaded 
on the bottom flange as shown. 




STRESSES 
CL=1H=EA=0; LD=32.5tons; 
EF=HG = 24t.ons; 
FO = GA= FG = DA =17.5 tons. 
KL^ 46 tons; KJ= 2.5 tons; 
DK = JB = 31.5 tons; 
JI = 4tons: DI = DH= 34.5 tons. 



Fig. 104. 



For convenience the frame is lettered in this instance in con- 
traclockwise order and the forces at the joints will be considered 
in like manner. 

To a scale of 4V inch = i ton, set off ah and he to represent 
in magnitude and direction the loads of 20 tons and 30 tons 
respectively. 



FRAMED STRUCTURES — RECIPROCAL DIAGRAM 165 

Selecting a pole we obtain, by means of the funicular polygon, 
cd and da for the reactions R2 and Ri respectively. 

For the reason already given there can be no stress in the 
members CL and EA, and this is indicated by placing / in the 
reciprocal diagram at the same point as c, and e at the same 
point as a. 

At the joint at the middle of the top flange there is a state of 
equilibrium under the action of the stresses in the members ID, 
DH, and HI, HI being at right angles to each of the others. 
There cannot, therefore, be any stress in HI, and h and i will 
fall at the same point in the reciprocal diagram. 

The members CL and EA add rigidity to the frame, and HI 
resists the tendency of the top flange to bend. 

For the equilibrium of the joint LDK we get the triangle Idk. 
For the joint CLKJB we get the force polygon clkjh, and so on 
to the completion of the reciprocal diagram. 

75. The Fink Truss. — The Fink truss of four bays. Fig. 105, 
has a span of 64 feet, a depth of 12 feet, and is uniformly loaded 
with 1.5 tons per foot, making 96 tons in all. The figure is 
constructed to a scale of 2V inch to the foot, and the load scale 
is taken as ^^ inch to the ton. The apportionment of the load 
places 12 tons over each support and they are rejected. 

In the construction of the reciprocal diagram there are in- 
sufficient data to begin at either of the support joints. A con- 
sideration of the equilibrium of joint 2 shows the stress in FI to 
be 24 tons. At joint 11 we have four forces acting along two 
lines; therefore the conditions of equilibrium require that the 
stress in FI shall equal that in HG, and that the stress in GF 
shall equal that in IH. Then the stress in HG is also 24 tons. 
At joint 10 we have equilibrium under the action of the stresses 
in GH, HE, and EG; and since HE and EG are equally inclined 
to GH their stresses must be equal. If, then, the stress in HE 
can be found, the stress in EG will be known, and there will be 



i66 



THE ELEMENTS OF MECHANICS OF MATERIALS 



sufficient data to begin at the joint at the left support. Since 
GH is a strut it is evident that the equihbrium of joint lo re- 
quires that HE and EG be ties, and therefore the direction of 
the stress in HE at joint lo, and also the direction of the stress 
in EG Sit joint i, are known. By means of the section xy and 
moments about the left support we have 

Stress in HE X 32 sin^ = 24 X 16, 
whence, Stress in HE = stress in EG = 20 tons. 




seitons 




/I 


r=^^. 




X^ 




\ 


L^^-" 



Fig. 105. 



There are now sufficient data to begin the construction of the 
reciprocal diagram by a consideration of the equilibrium of the 
joint at the left support. 

Set off the load line ad, making ab, be, and cd each equal to 
1^ inch in length to represent the external loads of 24 tons. 
Then de and ea are the right and left reactions respectively, to 
scale. 



FRAMED STRUCTURES — RECIPROCAL DIAGRAM 167 

From e draw eg parallel to EG, and make it %% inch long to 
represent the stress of 20 tons in EG. Commencing at e, we get 
eajg as the stress polygon for the joint at the left support. Know- 
ing a/, the stress polygon abifiov the joint ABIF is readily drawn, 
and so on to the completion of the reciprocal diagram. 

By means of the section ut and moments about the upper 
middle joint, we obtain 

Stress in JE X 32 sin o: + 24 X 16 = 36 X 32, 

whence 

Stress in JE = (36 X 32 - 24 X 16) v^ ^ 

96 ^^ 

which agrees with the scale length oije of the reciprocal diagram. 
The stresses in AF, BI, CL, and DO are shown by the diagram 
to be 80 tons each. This may be checked by the sections ut 
and vz. 

Stress in ^/ X 12 = 36 X 16 + stress in JE X 6 cos a, 
whence 

Stress in i5/ = 48 + 68.35 X -^ = 80 tons. 

V73 
From the section vz and moments about the joint FIHG we 
have 
Stress in EG X 2 V73 sin {§ — a) -\- stress in AF X 6 =36 X 16. 

whence. Stress in ^F = — — — ^ = 80 tons. 

6 

It will be noted that, the Fink truss is composed of a primary 
truss I, 8, 5, and two secondary trusses i, 10, 3, and 3, 6, 5. 
Whether or not there are joints at 11, 9, 7, and 12 will not affect 
the stresses in GF, FI, IE, HG, IJ, JE, KL, LM, ME, EK, 
LO, ON, NM. 

76. Roof Truss Fixed at the Ends and with Wind Pressure. — 
The roof truss of Fig. 106 is fixed at the ends and has the follow- 
ing data: 



THE ELEMENTS OF MECHANICS OF MATERIALS 




Fig. io6. 



FRAMED STRUCTURES — RECIPROCAL DIAGRAM 1 69 

Linear scale, i inch = 10 feet; load scale, 0.75 inch = i ton. 
Pitch of roof, 30°. Span of roof, 30 feet. Distance between 
trusses, 10 feet. Dead load per square foot of horizontal sur- 
face, 23 pounds. Horizontal wind pressure per square foot, 
40 pounds. 

The dead vertical load = ^ = 3 tons, approximately, 

2240 

and its uniform distribution places 0.75 ton at each of the 

joints 2,3, and 4, and 0.375 ton at each of the joints i and 5. 

In this case of a trass fixed at the ends, the effect of the wind 
and of the vertical loads at the end joints is borne entirely by 
the supports and does not affect the stresses in the members; 
the loads at the supports are therefore omitted from consider- 
ation, and the reactions to be used are those due only to the loads 
producing the stresses. 

The effective support reactions due to the vertical loads are 

each '-^ = 1. 125 tons = 1.125 X 0.75 = f|- inch when 

2 

reduced to scale. These reactions are vertical and are denoted 
in the frame diagram by R^. 

The vertical loads at the joints 2, 3, and 4 are, when reduced 
to scale, each equal to 0.75 X 0.75 = j^g inch. These vertical 
loads are laid off to scale at 'the joints. 

Suppose the wind to act on the left side of the roof. The 
pitch of the roof being 30°, and the horizontal intensity of 
the wind pressure being 40 pounds per square foot, the normal 
pressure is 26 pounds per square foot (see table of Art. 68). 

The total wind pressure to be borne by one truss is 

15 sec 30° X 10 X 26 , , 

-^ = 2 tons, very nearly. 

2240 

The distribution of this wind load places i ton at joint 2 and 
i ton at joint 3, the half ton at the left support being rejected. 
These loads to scale are J inch and f inch respectively, and are 



lyo THE ELEMENTS OF MECHANICS OF MATERIALS 

combined with the vertical loads so as to obtain the magnitudes 
and directions of the resultant pressures at the joints. 

The truss being fixed at the ends the support reactions due to 
the wind pressure must be parallel to and opposite in direction 
to the normal wind pressure in order that there shall be a balance 
between the external forces. The magnitude of these effective 
wind reactions can be found by means of a funicular polygon. 
Thus: 

Lay off the load Hne a'c' parallel to the normal wind pressure, 
making a^h' equal to the wind pressure at joint 2, and h'c' equal 
to the wind pressure at joint 3. Select at random a pole 0^ and 
draw the vectors oa! , oh' , and oc\ From some point 0' in the 
Hne of action of the left support reaction construct the funicular 
of the force diagram oa'c' . Draw oe' parallel to the closing line 
e'^o' of the funicular. Then, e'a' is the magnitude of the left 
support reaction RJ due to the wind, and c'e' is the magnitude 
of the right support reaction R,J' due to the wind. The re- 
sultant Ri of R^ and Rj, and the resultant R2 of R^ and Rj' 
are the total support reactions in magnitude and direction due 
to the dead load and the wind pressure. 

Having the magnitudes and directions of the resultant forces 
at all the joints, the reciprocal diagram can now be drawn. 
Thus: 

Draw ah, he, and cd parallel and equal to the resultant forces 
at joints 2,3, and 4. From d draw de parallel and equal to R2. 
The system being in equilibrium the force polygon must close; 
therefore the space ea must be exactly filled by a line parallel 
and equal to Ri\ otherwise the construction would be in- 
accurate. 

Commencing at the joint at the left support its reciprocal eafe 
is readily constructed. Proceeding then to joints 2, 3, 4, and 5 
the stresses in all the members are obtained, the necessity of 
the point i falling on fe affording a check as to accuracy. 



FRAMED STRUCTURES — RECIPROCAL DIAGRAM 



171 



The support reactions due to the wind might have been found 
easily by moments. The frame diagram being a scale drawing, 
the distances between the lines of action of the normal wind 
forces and of the reactions due to them are found by measure- 
ment to be as shown in Fig. 106. Then, taking moments about 
the right support, we have 

26 RJ = iXi.7+iX 0.86, whence RJ = 0.82 ton; 
hence, i^' = 0.82 X 0.75 = 0.615 inch = eV. 

Similarly, by moments about the left support, we have 

26 i?J' = I X 0.9 + 4 X 1.74, whence R^'' = 0.68 ton; 
hence, R^^' = 0.68 X 0.75 = 0.51 inch = cV. 

The stresses in the members, by scale measurements from the 
reciprocal diagram, are found to be as here tabulated: 



Ties 




Struts 



The solution of this problem with the consideration of all the 
forces, as shown in Fig. 107, rejecting none at the supports, is 
instructive. 

Lay off the load line abcdefghi to scale. Each of the reactions 
R^ is equal to 1.5 tons = 1X4=1 inch to scale. The effective 
wind reactions are found by the funicular as before, but the left 
reaction thus found must be augmented by the wind load AB 



172 THE ELEMENTS OF MECHANICS OF MATERIALS 




Fig. 107. 



FRAMED STRUCTURES — RECIPROCAL DIAGRAM 173 

in order to obtain the full reaction RJ at the left support due 
to the wind. Commencing at i the reactions RJ' , R^, RJ, R^ 
are each set off in magnitude and parallel direction, closing the 
force polygon at a. 

The reciprocal diagram can now be drawn, the stresses ob- 
tained being exactly the same as by the preceding method. The 
dotted lines aj and ji are the resultant support reactions Ri and 
R2 in magnitude and parallel direction. 

77. Roof Truss Fixed at One End and Free at the Other with 
Wind Pressure and Dead Vertical Loads. — The roof truss of 
Fig. 108 has a span of 50 feet, and is fixed at one end and free 
to move on expansion rollers at the other. Pitch of roof, 28° ; 
distance between trusses, 16 feet; dead load per square foot of 
horizontal surface, 17.9 pounds; wind pressure normal to the 
truss, 19.77 pounds per square foot. Linear scale, tV inch = i 
foot; load scale, ^ inch = i ton. The frame diagram shows the 
angular arrangement of the truss members, and it will be noticed 
that the joints divide the rafters into three parts whose lengths 
are to each other as 3 : 3 : 2. 

The dead vertical load on each truss = ^^ '—^ = 6.4 

2240 

tons, 3.2 tons on each rafter, ^and its uniform distribution places 
0.6 ton at joints i and 7, 1.2 tons at joints 2 and 6, i ton at joints 
3 and 5, and 0.8 ton at joint 4. 

The support reactions due to the vertical loads are each 

6 4. ^2 I '\2 

—^ = 3.2 tons = — X - = ^— inches when reduced to scale. 
2 10 2 20 

These reactions are vertical, and each is denoted by R^ in the 

frame diagram. The scale measurements of the vertical loads 

at the joints are: 0.6 X 0.5 = 0.3 inch at joints i and 7; 

1.2 X 0.5 = 0.6 inch at joints 2 and 6; 0.5 inch at joints 3 and 5; 

and 0.4 inch at joint 4. These loads are laid off to scale at the 

joints in the frame diagram. 




rig.109 
Scale, J4' — 1 ton 



(174) 



FRAMED STRUCTURES — RECIPROCAL DIAGRAM 175 

The total wind pressure to be borne by one truss is 

2% sec 28° X 16 X 1Q.77 

— ^-^ = 4 tons. 

2240 

This load may act on either side of the truss. Suppose first 
that the left end of the truss be fixed and the wind to act on that 
side. The distribution of the wind load places 0.75 ton at 
joint I, 1.5 tons at joint 2, 1.25 tons at joint 3, and 0.5 ton at 
joint 4. Reduced to scale these loads are: At joint i, 0.75 
X 0.5 = 0.375 = 8 inch; at joint 2, 1.5 X 0.5 = | inch; at joint 
3, 1.25 X 0.5 = 0.625 = f inch; and at joint 4, 0.5 X 0.5 = 0.25 
= J inch. Set off these loads to scale at the joints in the frame 
diagram. 

The right end being free, the reaction there due to the wind 
pressure will be vertical. 

Assume the truss to be acted on only by the wind and that its 
resultant pressure of 4 tons acts at the middle point of the left 
rafter. There are now but three concurrent forces, viz., the 
two support reactions and the resultant wind pressure. The 
points of appHcation of these forces and the directions of two 
of them being known, the direction of the third can be found. 
Thus : 

Let fall a perpendicular to the left rafter from its middle 
point, and at the right support let fall a perpendicular to the 
bottom chord of the truss. These perpendiculars are the di- 
rections of the resultant normal wind pressure and of the 
support reaction at the free end respectively, and they intersect 
at X. The line joining x with joint i will give the direction of 
the support reaction due to the wind at the left (fixed) end. Lay 
off to scale the distance xz equal to the magnitude, 4 tons, of 
the resultant normal wind pressure, and resolve it into its com- 
ponents xy and yz parallel to the support reactions. The reac- 
tion Rj at the left support due to the wind is represented in 



176 THE ELEMENTS OF MECHANICS OF MATERIALS 

magnitude by xy. The composition of Rj and R^ gives the 
resultant reaction R\ at the fixed end due to all the external 
forces. The addition of yz to R^ gives R^ as the support reac- 
tion at the right (free) end due to all the external forces. The 
resultants of the dead and wind loads at joints i, 2, 3, and 4 are 
found by the parallelogram of forces as shown. 

The reciprocal diagram for the truss with its left end fixed 
and the wind blowing on the fixed side can now be drawn. 

To a scale of 0.25 inch to the ton — one-half that used for 
the loads in the frame diagram — construct the force polygon 
ahcdefghia of Fig. 109. This polygon must, of course, close. 
The diagram is readily completed by drawing in order the 
reciprocals of the joints lABJ, JBCK, IJKL, LKCDM, 
MDEN, ILMNO, ONEFP, and PFGQ, the point q falling at 
the finish on the line ij. The order of procedure from joint 
to joint determines itself from the fact that the reciprocal of 
a joint cannot be drawn if more than two of the forces are 
unknown. 

It will be necessary to construct the reciprocal diagram for 
the wind blowing on the free side and, since the vertical loads 
are the same on each side, the construction will be facilitated by 
assuming, in Fig. 108, the right end of the truss to be fixed and the 
left end free to move on rollers. 

The reaction at the left end will now be vertical, and its Hne 
of action intersects the line of action of the resultant normal 
wind pressure on the free side at x\ and therefore the line 
joining x^ with joint 7 gives the direction of the reaction at the 
right support due to the wind. Lay off x^z^ equal to 4 tons to 
scale to represent the magnitude of the resultant wind pressure. 
Its component x'y' gives the magnitude of RJ', the reaction at 
the fixed end due to the wind. The composition of Rj^ and R^ 
gives R2 as the support reaction at the right (fixed) end due to 
all the loads. The component y^z^ is the amount to be added to 



FRAMED STRUCTURES — RECIPROCAL DIAGRAM 



177 



R„ at the left support to give the total support reaction Ri at 
the free end. 

The force polygon and the reciprocal diagram can now be 
drawn as shown in Fig. no, the point q falling on the line ij. 

The members must be designed to resist the maximum stress 
to which they may be subjected, and since the wind may blow 
on either side of the truss they are made the same for each side. 
The tabulated stresses were found by scale measurements 
from Figs. 109 and no to be as follows: 

TABLE OF STRESSES. 



Members. 


Wind on fixed side. 


Wind on free side. 


Ties. 


Struts. 


Ties. 


Struts. 


BJ 




9.20 




9-25 


JI 


9.60 


7.80 


CK 


8.50 
2.55 


8.50 
2 60 


KJ 






KL 


2.5s 
6.90 


2.50 
5-05 




LI 






DM 


5.60 
3-40 
5.80 


5 -60 
3 40 
5-8o 


ML 






EN 






NM 


4-15 


4-05 


NO 


1-45 


1-35 


01 


5-90 


4.00 


FP 


7-30 


7.35 


PO 


1. 00 ' 


1. 10 


GQ 
QP 

Qi 


7-95 
1. 00 


8 00 






1 . 10 


7.00 


5.20 









78. A Framed Crane. — To draw the reciprocal diagram of 
the framed crane of Fig. in, having a load of 3 tons at its peak, 
we proceed as follows: 

Letter the frame diagram, and then to a scale of 0.25 inch = 
I ton lay off the load line ab three-fourths of an inch in length to 
represent the load of 3 tons. From b and a draw parallels to 
BJ and J A respectively. They intersect atj, giving abj as the 
force diagram of the joint ABJ, and bj and ja represent in 



178 



THE ELEMENTS OF MECHANICS OF MATERIALS 



magnitude and direction the stresses in the members BJ and J A 
respectively. Proceed in Hke manner to the completion of the 







reciprocal diagram from which the stress diagram for any joint 
may at once be read. Thus, the stress diagram for the joint 
GHBF is ghbfg. The reaction R2 is, of course, equal to Ri + W. 



FRAMED STRUCTURES — RECIPROCAL DIAGRAM 179 

79. Redundant and Deficient Frames. — The reciprocal dia- 
grams for all the frames that have been considered have closed, 
and the frames, therefore, were complete. It will be noted that 
in the cantilever frames the number of members equals twice 
the number of joints minus four, and that in frames supported 
at the ends the number of members equals twice the number of 
joints minus three. 

If a frame has more than the requisite number of members to 
retain its original shape, it is called a redundant frame and is 
statically indeterminate. 

A frame having an insufficient number of members to pre- 
serve its original shape is deficient, but may be used for one 
special distribution of the load. Any deviation from this dis- 
tribution will cause the frame to change its shape. A deficient 
frame may generally be made complete by the addition of another 
member. 



PROBLEMS 

I. Draw the reciprocal diagram of the Warren girder of Fig. c, the 
triangles being equilateral. Span, 42 feet, and a uniformly distributed 
load of 1.5 tons per foot is sustained by the girder. Tabulate the stresses 
and indicate their kinds in the frame diagram. Scales: Load, o.i inch = 
I ton; linear, o.i inch = i foot. 




2. Draw the reciprocal diagram of the roof truss of Fig. d, which is 
loaded as shown. Tabulate the stresses and indicate their kinds in the 
frame diagram. Scales: Load, 0.25 inch = i ton; linear, 0.1 inch = i 
foot. 



i8o 



THE ELEMENTS OF MECHANICS OF MATERIALS 



3 tons 



4 tons 




Fig. d. 



3. The Fink truss of Fig. e has a span of 56 feet, a depth of 12 feet, 
and is uniformly loaded with 1.5 tons per foot. Draw the reciprocal dia- 
gram to the scale of i^ inch = i ton. Tabulate the stresses and indicate 
their kinds in the frame diagram. Scale: Linear, ^ inch = i foot. 



A 




B 






CD 


^>^. 




1 


/ 


\ 


s L 


'-^^ 


^ 


^ 


< 


J 


K 


>^ 


V' 



Fig. e. 



4. The braced cantilever of Fig. / is 20 feet long, 9 feet deep, and uni- 
formly loaded on the top with 120 pounds per foot. Draw the reciprocal 
diagram to the scale of 0,1 inch = 100 pounds. Tabulate the stresses and 
indicate their kinds in the frame diagram. Show that the reactions at the 
wall are equal but opposite in direction. Scale: Linear, te inch = i foot. 




Fig. /. 



5. Draw the reciprocal diagram of the braced cantilever of Fig. g, and 
tabulate the stresses. Find, in magnitude and direction, the resultant 
stress on the pin of the upper joint at the wall. Scales: Load, o.i inch 
= 80 pounds; linear, 0.25 inch = i foot. 



FRAMED STRUCTURES — RECIPROCAL DIAGRAM l8l 

^' 200o[Lbs, 




6. A roof truss, Fig. h, of 50 feet span, fixed at the ends and of 30° 
pitch, carries vertical loads of 1.5 tons at AB, BC, CD, DE and EF, and 
sustains a horizontal wind pressure of 33.5 pounds per square foot. Dis- 
tance between principals, 12 feet. Draw the reciprocal diagram and 
tabulate the stresses in the members. Scales: Linear, ^ inch = i foot; 
load, \ inch = i ton. 




Fig. h. 

7. The roof truss of Fig. i has a span of 75 feet and a pitch of 30°. 
The rafters are divided into three equal parts and the horizontal tie into 
five equal parts. The distance between principals is 16 feet. The dead 
vertical load is 6.5 tons, distributed as follows: At joints i and 7, 0.6 ton 




l82 



THE ELEMENTS OF MECHANICS OF MATERIALS 



each; at joint 4, i ton; at joints 2 and 3, 0.95 ton each; at joints 5 and 6, 
1.2 tons each. The wind pressure is assumed to have a horizontal inten- 
sity of 20 pounds per square foot. Assuming the truss to be fixed at one 
end and free to move on expansion rollers at the other, determine the 
maximum stress in each member. Scales: Linear, -^q inch =1 foot; load, 
i inch = I ton. 

8. The roof truss of Fig. j has a span of 60 feet, and supports dead 
loads of 2 tons at each of the joints 2, 3, 4, 5, 6, 7, 8. Draw its reciprocal 
diagram. Scales: Linear, xa inch = i foot; load, ^ inch = i ton. 




Fig. j. 

9. A bridge truss, a portion of which is shown in Fig. k, has a span of 
120 feet divided into six equal bays. It is designed to carry a dead load 
of 16 tons at each of the lower panel points. Calculate, by the method of 
sections, the stresses in the members GM, MN, and NC, and indicate by 
arrowheads the nature of the stresses. 




Fig. k. 



framed' structures — RECIPROCAL DIAGRAM 



183 



10. Complete the frame diagram of problem 9 and construct the re- 
ciprocal diagram, tabulating the magnitudes and kinds of stresses in the 
members. Scales: Linear, 



jV inch = I foot; load, ^\ inch = i ton. 



II. The crane of Fig. / supports a load of 2 tons at its peak. Deter- 
mine from the reciprocal diagram the reactions and the stresses in the 
members. Scales: Linear, i inch = i foot; load, i inch = i ton. 




2 tons 



Fig. /. 

12. Calculate by the method of sections the magnitude and kind of 
stress in each of the members of the crane of Fig. m that is cut by the hne xy. 




Fig. 7)1. 

13. Draw the reciprocal diagram of the crane of Fig. m. Scales: 
Linear, y^ inch = i foot; load, jV iiich = i ton. 



CHAPTER rX 
ENGINEERING MATERIALS 

80. Introductory. — In all engineering construction a knowl- 
edge of the properties of the materials used is essential. It is 
not sufficient to know only the characteristic properties of 
different materials, but their fitness for use in any engineering 
design must be determined by actual test in order to insure 
safety to the structure, and for this purpose the modern 
mechanical laboratory is equipped with special machines for 
testing. 

The materials used in machine construction must be strong 
and practically rigid, the element of strength enabling the size 
and weight of machines to be reduced to a minimum, and the 
rigidity enabling the material to resist the tendency to change 
its size or shape while under the influence of straining actions. 
As no materials are absolutely rigid, it follows that the materials 
of machines undergo some deform^ation under the action of the 
forces to which they are subjected, but by judicious selection of 
materials and a proper proportioning of parts, such deforma- 
tions may be rendered so small as to be negligible. Generally 
speaking, the deformations in machines are temporary in char- 
acter, the elasticity of the chosen materials causing them to 
vanish upon the removal of the straining force. 

The principal engineering materials are cast iron, wrought iron, 
steel, copper, lead, tin, zinc, alloys, timber, concrete, and reinforced 
concrete. 

81. Cast iron. — Cast iron is obtained in the form of pig 
iron by direct treatment of the ore in a blast furnace, its quality 

184 



ENGINEERING MATERIALS 185 

depending on the relative amounts of other substances in its 
combination. 

Influence of Carbon. — In the molten condition the carbon is 
dissolved by the iron and held in solution just as ordinary salt 
is ^dissolved by water. The mixture or combination of the two 
. elements is thus entirely uniform. The proportion of carbon 
which pure melted iron can thus dissolve and hold in solution 
is about 3.5 per cent. If chromium or manganese is present also, 
the capacity for carbon is much increased, while with silicon, on 
the other hand, the capacity for carbon is decreased. In the 
various grades of cast iron the proportion of carbon is usually 
found to be between 2 per cent and 4.5 per cent. 

When such a molten mixture cools and becomes solid there is 
a tendency for a part of the carbon to be separated and no longer 
remain in intimate combination with the iron. The carbon thus 
separated, or precipitated, from the iron takes that form known 
as graphite and collects in very small flakes or scales. The 
carbon v/hich remains in intimate combination with the iron is 
said to be combined, while that which is separated is usually 
called graphitic. 

The qualities of cast iron depend chiefly on the proportion of 
total carbon and on the relative proportion of combined and 
graphitic carbon. 

With a high proportion of graphitic carbon the iron is soft and 
tough, with a low tensile strength, and breaks with a dark and 
coarse-grained fracture. In fact, the substance in this condition 
may be considered as nearly pure iron with fine flakes of graphite 
entangled and distributed through it, thus giving to the iron a 
spongy structure. The iron thus forms a kind of continuous 
mesh about the graphite, which decreases the strength by reason 
of the decrease of cross-sectional area actually occupied by the 
iron itself. Such irons are termed gray. 

As the relative proportion of graphitic carbon decreases and 



1 86 THE ELEMENTS OF MECHANICS OF MATERIALS 

that of combined carbon increases, the iron takes on new prop- 
erties, becoming harder and more brittle. Its tensile strength 
also increases to a certain extent, and the fracture becomes fine- 
grained, or smooth, and whiter in color. When these charac- 
teristics are pronounced, the iron is said to be white. When 
about half the carbon is combined and half separates as graphite 
the effect is to produce a distribution of dark spots scattered 
over a whitish field. Such iron is said to be mottled. 

In a general way, with a large proportion of total carbon 
there is likely to be formed a considerable amount of graphitic 
carbon, and hence such iron is usually gray and soft. Also, 
with a large proportion of carbon the iron melts more readily 
and its fluidity is more pronounced. As the proportion of total 
carbon decreases the cast iron gradually approaches the con- 
dition of steel, whose properties will be discussed in later 
paragraphs. 

Of the special ingredients in cast iron the combined carbon is 
the one of greatest importance. It is that chiefly which, by unit- 
ing with the iron, gives it new qualities, and the principal influ- 
ence of other substances lies in the effect which they may have 
on the proportion of this ingredient. As between graphitic 
and combined carbon the former does not affect the quality 
of the iron itself, but acts physically by affecting the structure 
of the casting, while the latter by entering into combination with 
the iron acts chemically, and produces a new substance with 
different qualities. 

The proportions of combined and graphitic carbon are influ- 
enced by the rate of cooling and by the presence or absence of 
various other ingredients. Slow cooling allows time for the 
separation of the carbon and thus tends to form graphitic carbon 
and soft gray irons. Quick cooling, or chilling in the extreme 
case, prevents the formation of graphitic carbon and thus tends 
to form hard, white iron. 



ENGINEERING MATERIALS 187 

In addition to carbon, small particles of silicon, sulphur, phos- 
phorus, manganese, and chromium may be found in cast iron. 

Influence of Silicon. — The fundamental influences of silicon 
are two. (a) It tends to expel the carbon from the combined 
state and thus to decrease the relative proportion of combined 
carbon and increase that of graphitic carbon, (b) Of itself, 
silicon tends to harden cast iron and to make it brittle. These 
two influences are opposite in character, since an increase in 
graphitic carbon softens the iron. In usual cases the net result 
is a softening of the iron, an increase in fluidity, and a general 
change toward those qualities possessed by iron with a high 
proportion of graphitic carbon. This appHes with a proportion 
of silicon from 2 per cent to 4 per cent. With more than this 
the influence on the carbon is but slight and the result on the 
iron is to decrease the strength and toughness, giving a hard but 
brittle and weak grade of iron. 

A chilled cast iron is an iron which, if cooled slowly, would be 
giay and soft, but if cooled suddenly by contact with a metal 
mold, or by other means, becomes white or hard, especially at 
and near the surface. Certain grades of cast iron tend to chill 
when cast in sand molds. This property is usually undesirable. 
In such cases the tendency may be prevented by the addition of 
silicon, which, by forcing the carbon into the graphitic state on 
cooling, prevents the formation of hard, chilled surfaces. In all 
cases the actual effect of adding silicon will depend much on the 
character of the iron used as a base, and only a statement of the 
general tendencies can here be given. 

To sum up, a white iron which would give hard, brittle, and 
porous castings can be made softer, tougher, and more solid by 
the addition of silicon to the extent of perhaps 2 per cent or 3 
per cent. As the siKcon is increased, the iron will become softer 
and grayer and the tensile strength will decrease. At the same 
time the shrinkage will decrease, at least for a time, though it 



1 88 THE ELEMENTS OF MECHANICS OF MATERIALS 

may increase again with large excess of silicon. The softening 
and toughening influence, however, will only continue so long as 
additional graphite is formed, and when most of the carbon is 
brought into this state the maximum effect has been produced 
and any further addition of silicon will decrease both strength 
and toughness. 

Influence of Sulphur. — Authorities are not in entire agree- 
ment as to the influence of sulphur on cast iron, some beheving 
that it tends to increase the proportion of combined carbon, 
while others maintain that it tends to decrease both the com- 
bined carbon and the silicon. It is generally agreed, however, 
that in proportions greater than about 0.15 to 0.20 of i per 
cent it increases the shrinkage and the tendency to chill and 
decreases the strength. Sulphur does not readily enter cast iron 
under ordinary conditions and its influence is not especially 
feared. An increase in the proportion of sulphur in cast iron is 
most likely to result from an absorption of sulphur in the coke 
during the operation of melting in the cupola. 

Influence of Phosphorus. — The presence of phosphorus in- 
creases the fusibility, fluidity, and brittleness of cast iron and is 
desirable in Hght and ornamental castings. The maximum 
amount of phosphorus should not exceed 1.5 per cent. 

Influence of Manganese. — This element by itself decreases 
fluidity, increases shrinkage, and makes the iron harder and more 
brittle. It combines with iron in all proportions. The com- 
bination containing less than 50 per cent of manganese is called 
spiegeleisen. With more than 50 per cent of manganese it is 
called ferromanganese. One of the most important properties of 
manganese in combination with iron is that it increases the 
capacity of the iron for carbon. Pure iron will take only about 
3.5 per cent of carbon, while with the addition of manganese the 
proportion may rise to 6 or 7 per cent. Manganese is also be- 
lieved to decrease the capacity of iron for sulphur and to this 



ENGINEERING MATERIALS 189 

extent may be a desirable ingredient in proportions not exceed- 
ing I to 1.5 per cent. 

Shrinkage. — At the moment of hardening, cast iron expands 
and takes a good impression of the mold. In the gradual cooling 
after setting, however, the metal contracts, so that on the whole 
there is a shrinkage of about 0.125 inch per foot in all directions, 
though this amount varies somewhat with the quality of the 
iron and with the form and dimensions of the pattern. In a 
general way, hardness and shrinkage increase and decrease 
together. 

Strength and Hardness. — The strength of cast iron is chiefly 
dependent upon its amount of combined carbon. The greatest 
crushing strength is obtained with sufficient combined carbon 
to make a rather hard, white iron, while for the maximum 
transverse or bending strength the combined carbon is some- 
what less and the iron only moderately hard. For the greatest 
tensile strength, the combined carbon is still less and the iron 
rather soft. Metal still softer than this works with the greatest 
facility but is deficient in strength. 

Uses in Engineering. — Cast iron is used for cyhnders, cylin- 
der heads, liners, shde valves, valve chests and connections, and 
generally for all parts having considerable complexity of form. 
It is also used for columns, bed plates, bearing pedestals, caps, 
etc., though cast and forged steel are to some extent displacing 
cast iron for some of these uses. It is also used for grate bars, 
furnace door frames, and for minor boiler fittings. 

Inspection of Castings. — In the inspection of castings care 
must be had to note the texture of the surface and to this end 
the outer scale and burnt sand should be carefully removed by 
the use of brushes or chipping hammer, or, if necessary, by 
pickhng in dilute muriatic acid. The flaws most liable to occur 
are blow holes and shrinkage cracks. The parts of the casting 
most liable to be affected by blow holes are those on the upper 



IQO THE ELEMENTS OF MECHANICS OF MATERIALS 

side or near the top, and on this account a sinking head or extra 
piece is often cast on top, into which the gases and impurities 
may collect. This is afterwards cut off, leaving the sounder 
metal below. Shrinkage cracks are of unusual occurrence. The 
presence of blow holes, if large in size or in great number and 
near the surface, may often be determined by tapping with a 
hammer. The sound given out will serve to indicate to an ex- 
perienced ear the probable character of the metal underneath. 

Brazing of Cast Iron. — Cast iron may be brazed to itself, or 
to most of the structural metals, by the use of a brazing solder 
of suitable melting point and with proper care in the operation. 
Cast iron may also be united to itself, or to wrought iron or 
steel, by the operation of burning. This consists in placing in 
position the two pieces to be united, and then allowing a stream 
of melted cast iron to flow over the surfaces to be joined, the 
adjacent parts being protected by fire clay or other suitable 
material. The result is to soften or partially melt the surfaces 
of the pieces, and by arresting the operation at the right moment 
they may be securely joined together. 

Malleable Cast Iron. — If iron castings of not too great thick- 
ness, and of such purity as to be low in sulphur, be embedded in 
powdered red oxide of iron (red hematite) and maintained at a 
red heat for two or three days, they become, in a measure, de- 
carbonized as a result of the chemical action which ensues. 
The carbon first disappears from the outer layers, and as the 
process continues decarbonization toward the innermost layers 
takes place. If the process be carried to the extreme in the 
effort to withdraw all the carbon from the interior the outer 
layer is very liable to become brittle, thus defeating the object 
to be attained. For this reason there always remains a core of 
cast iron only partially decarbonized, while the outermost layers 
are left in the condition of soft or malleable iron. The process 
has little effect upon the sulphur, manganese, phosphorus, and 



ENGINEERING MATERIALS IQI 

Other impurities of the castings, but the resulting product is a 
malleable casting much less fusible than cast iron and possessing 
six times its ductility. The best product may be twisted and 
bent to a considerable extent before breaking, and its ability to 
withstand shocks is much greater than that of cast iron. Pipe 
fittings, to some extent, are malleable castings, and this mate- 
rial is largely used in appliances of car construction where more 
strength and toughness are required than cast iron affords. 

82. Wrought Iron. — Wrought iron is nearly pure iron 
mixed with more or less slag. Nearly all the wrought iron used 
in modern times is made from cast iron by the puddling process 
in a reverberatory furnace. For the details of this process 
reference may be had to textbooks on metallurgy. We can only 
note here that, in a furnace somewhat similar to the open hearth, 
most of the carbon, silicon, and other special ingredients of 
cast iron are removed by the combined action of the flame and 
of a molten bath of slag or fluxing material, consisting chiefly of 
black oxide of iron. As this process approaches completion 
small bits of nearly pure iron separate from the bath of melted 
slag and unite. This operation is assisted by the puddling bar, 
and after the iron has thus become separated from the liquid slag 
it is taken out, hammered or squeezed, and rolled into bars or 
plates. Some of the slag is necessarily retained in the iron and, 
by the process of manufacture, is drawn out into fine threads, 
giving to the iron a stringy or fibrous appearance when nicked 
and bent over or pulled apart. 

The proportion of carbon in wrought iron is very small, rang- 
ing from 0.02 to 0.20 of I per cent. In addition, small amounts 
of sulphur, phosphorus, silicon, and manganese are usually 
present. 

The proportion of sulphur should not exceed o.oi of i per cent. 
Excess of sulphur makes the iron red-short, that is, brittle when 
red hot. 



192 THE ELEMENTS OF MECHANICS OF MATERIALS 

The proportion of phosphorus may vary from 0.05 to 0.25 of 
I per cent. Excess of phosphorus makes the metal cold-short, 
that is, brittle when cold. 

The proportion of sihcon may vary from 0.05 to 0.30 of i per 
cent. 

V The proportion of manganese may vary from 0.005 to 0.05 of 
1 per cent. The influence of the silicon and manganese is usu- 
ally slight and unimportant. 

Special Properties. — Wrought iron is malleable and ductile, 
and may be rolled, flanged, forged, and welded. It cannot be 
hardened, though by the process of case-hardening a surface 
layer of steel is formed which may be hardened. Wrought iron 
may be welded, because for a considerable range of temperature 
below melting (which takes place only at a very high tempera- 
ture) the iron becomes soft and plastic, and two pieces pressed 
together in this condition unite and form, on cooling, a junction 
nearly as strong as the solid metal. In order that this welding 
operation may be successful, the iron must be heated sufficiently 
to bring it to the plastic condition, yet not overheated, and 
there must be employed a flux (usually borax) which will unite 
with the iron oxide and other impurities at the joint and form a 
thin liquid slag which may readily be pressed out in the oper- 
ation, thus allowing the clean metal surfaces of the iron to effect 
a union as desired. 

83. Steel. — Steel may be made from wrought iron by in- 
creasing its proportion of carbon, or from cast iron by decreas- 
ing its proportion of carbon. The earlier processes followed the 
first method, and high-grade steels are still made in this way by 
the crucible process. 

The properties of steel depend partly on the proportions of 
carbon and other ingredients which it may contain, and partly 
on the process of manufacture. The proportion of carbon is 
intermediate between that for wrought iron and for cast iron. 



ENGINEERING MATERIALS 193 

In the so-called mild or structural steel the carbon is usually 
from 0.1 to 0.3 of I per cent. In spring steel the carbon propor- 
tion is somewhat greater, and in high carbon grades, such as are 
used for tools, etc., the carbon is from 0.6 to 1.2 per cent. In 
addition to the carbon there may be sulphur, phosphorus, silicon, 
and manganese in varying but very small amounts. 

Crucible Steel. — In the crucible process of making steel a 
pure grade of wrought iron is rolled into flat bars. These are 
then cut, piled, and packed with intermediate layers of charcoal 
and subjected to a high temperature for several days. This 
recarbonizes or adds carbon to the wrought iron and thus makes 
what is called cement or blister steel. These bars are then broken 
into pieces of convenient size, placed in small crucibles, melted, 
and cast into bars, or into such shapes as are desired. 

Structural Steel. — Structural or mild steel is made by the 
second general process, that of reducing the proportion of carbon 
in cast iron. In this operation there are two processes, known 
as the Bessemer and the Siemens-Martin or open hearth. 

Bessemer Process. — In this process the carbon and sihcon 
are burned almost entirely out of the cast iron by forcing an 
air blast through the molten iron in a vessel known as a con- 
verter. A small amount of spiegeleisen, or iron rich in carbon 
and manganese, is then added in such quantity as to make the 
proportion of carbon and manganese suitable for the charge as 
a whole. The steel thus formed is then cast into ingots, or into 
such forms as may be desired. In this process no sulphur or 
phosphorus is removed, so that it is necessary to use a cast iron 
nearly free from these ingredients in order that the steel may 
have the properties desired. 

A modification, by means of which the phosphorus is removed, 
and known as the basic Bessemer process, is used to some ex- 
tent. In this process, calcined or burnt lime is added to the 
charge just before pouring. This unites with the phosphorus, 



194 THE ELEMENTS OF MECHANICS OF MATERIALS 

removes it from the steel, and brings it into the slag. In the 
basic process the converter is lined with ganister — a mixture 
of ground quartz and fire clay — to protect it from attack by 
the limestone added to the charge and from the resulting slag. 

In the Bessemer process first noted, often known as the acid 
process, in distinction from the basic process, the lining of the 
converter is of ordinary fire clay. 

The removal of the phosphorus by the basic process makes 
possible the use of an inferior grade of cast iron. At the same 
time, engineers are not altogether agreed as to the relative 
values of the two products, and many prefer steel made by the 
acid process from an iron nearly free from phosphorus at the 
start. 

The Open-hearth Process. — In this process a charge of 
material consisting of wrought iron, cast iron, steel scrap, and 
sometimes certain ores, is melted on the hearth of a reverbera- 
tory furnace heated by gas fuel on the Siemens-Martin or regen- 
erative system. The carbon is thus partially burnt out in much 
the same manner as for wrought iron, and the proportion of 
carbon is brought down to the desired point, or slightly below 
that point. A charge of spiegeleisen is then added in order that 
the manganese may act on any oxide of iron slag which remains 
in the bath which, if allowed to form a part of the charge, would 
make the steel red-short. The manganese separates the iron 
from the oxide and returns it to the bath, while the carbon joins 
with that already present and thus produces the desired propor- 
tions. 

Here, as with the similar operations with the Bessemer con- 
verter, there is no removal of sulphur or of phosphorus, and only 
materials nearly free from these ingredients can be used for 
steel of satisfactory quality. With very low carbon, however, a 
little phosphorus seems to be desirable to add strength to the 
metal. This limitation of the available materials has led, as with 



ENGINEERING MATERIALS 1 95 

the Bessemer process, to the use of calcined limestone in the 
charge, its purpose being to unite with most of the phosphorus 
and hold it in the slag. As with the Bessemer process, it is 
necessary in this case to use a basic lining for the furnace, and 
it is known as the basic open-hearth process. The process 
which does not use limestone in the charge has come to be 
known as the acid open-hearth process. There is much differ- 
ence of opinion as to the relative merits of the two open-hearth 
processes. Either will produce good steel with proper care and 
neither will without it. It is usually considered sufhcient to 
specify the allowable limits of phosphorus and sulphur and 
then leave the choice of the acid or basic process to the 
maker. 

Open-hearth and Bessemer Steels Compared. — Open-hearth 
steel is usually preferred for structural material, for these reasons : 

(a) It seems to be more reliable and less subject to unexpected 
or unexplained failure than the Bessemer product. 

{h) Analysis shows that it is much more homogeneous in 
composition than Bessemer steel, and experience shows that it 
is much more uniform in physical quality. This is due to the 
process of manufacture, which is much more favorable to a 
thorough mixing of the charge than in the Bessemer process. 

{c) The open-hearth steel may be tested from time to time 
during the operation, so that its composition may be determined 
and adjusted to fulfill special conditions. This is not possible 
with the Bessemer process. 

Influence of Sulphur. — Sulphur makes steel red-short and 
interferes with its forging and welding properties. Manganese 
tends to counteract the bad effects of sulphur. Good crucible 
steel has rarely more than o.oi of i per cent of sulphur. In 
structural steel the proportion may vary from 0.02 to o.io of i 
per cent. When possible it should be reduced to not more than 
0.03 or 0.04 of I per cent. 



196 THE ELEMENTS OF MECHANICS OF MATERIALS 

Influence of Phosphorus. — Phosphorus increases the tensile 
strength and raises the elastic limit of low carbon or structural 
steel, but at the expense of its ductility and toughness, or ability 
to withstand shocks and irregularly applied loads. It is thus 
considered as a dangerous ingredient and the amount allowable 
should be carefully specified. This is usually placed from 0.02 
to o.io of I per cent. 

Influence of Silicon. — Silicon tends to increase the tensile 
strength and to reduce the ductility of steel. It also increases the 
soundness of castings and ingots and, by reducing the iron oxide, 
tends to prevent red-shortness. The process of manufacture 
usually removes nearly all of the silicon, so that it is not an ele- 
ment likely to give trouble to the steel makers. The proportion 
allowable should not be more than from o.i to 0.2 of i per 
cent. 

Influence of Manganese. — This element is believed to in- 
crease hardness and fluidity, and to raise the elastic limit and 
increase the tensile strength. It also removes iron oxide and 
sulphur and tends to counteract the influence of such amounts 
of sulphur and phosphorus as may remain. It is thus an im- 
portant factor in preventing red-shortness. The proportion 
needed to obtain these valuable effects is usually found between 
0.2 and 0.5 of I per cent. 

Semisteel. — A metal bearing this trade name has in recent 
years attracted favorable attention and has come into consid- 
erable use where somewhat greater strength and toughness are 
required than can be provided by cast iron. It is made by 
melting mild steel scrap, such as punchings and clippings of 
boiler plate, with cast-iron pig in the proportions of 25 or 30 
per cent of the former to 75 or 70 per cent of the latter. The 
presence of manganese and other special fluxes in small pro- 
portions is found to add essentially to the strength, toughness, 
and abihty to withstand shocks decidedly greater than for cast 



ENGINEERING MATERIALS 197 

iron, and with fairly good machine quahties. Semisteel casts 
as readily as most grades of cast iron, and its shrinkage and 
general manipulation are about the same. 

Mechanical Properties. — The tensile strength of the lowest 
carbon steel, say about o.io of i per cent carbon, is usually 
from 50,000 to 55,000 pounds per square inch of section. The 
strength increases quite uniformly with the increase of carbon, 
provided there are no unusual proportions of sulphur and phos- 
phorus. Experiment shows that under these circumstances the 
tensile strength will increase up to 75,000 pounds per square 
inch, or higher, at the rate of from 1200 to 1500 pounds per o.oi 
of I per cent of carbon added. At the same time, with the in- 
crease in strength the ductility decreases, so that a proper choice 
must be made according to the particular uses for which the 
steel is intended. In the best grades of tool steel, with carbon 
ranging from 0.5 to i.o per cent or over, the strength ranges 
from 80,000 to 120,000 pounds, and even higher in exceptional 
cases. 

Special Properties. — Mild or low carbon steel may be welded, 
forged, flanged, rolled, and cast. It cannot be tempered or 
hardened "with a proportion of carbon lower than about 0.75 of 
I per cent. High carbon -steel can be welded only imperfectly 
and, if very high in carbon, not at all. It can be forged with 
care and cast in forms as desired. It can be tempered or hard- 
ened by heating to a full yellow and quenching in cold water or by 
other means, and then drawing the temper to the point desired. 

Mild steel should not be worked under the hammer or flanging 
press at a low, or blue, heat, as such working is found in many 
cases to leave the metal brittle and unreliable. Steel, in order 
to weld satisfactorily, should have a low proportion of sulphur, 
and special care is required in the operation, because the range 
of temperature through which the metal is plastic and fit for 
welding is less than with wrought iron. 



igS THE ELEMENTS OF MECHANICS OF MATERIALS 

Tempering. — In the operation of tempering, the steel after 
quenching is very hard and brittle. In order to give the metal 
the properties desired, the temper is drawn down by reheating 
it to a certain temperature and then quenching again; or better 
still, by allowing it to cool gradually, provided the temper does 
not rise above the limiting value suitable for the purpose de- 
sired. If the reheating is done in a bath of oil, the conditions 
may be kept under good control and the final cooling may be 
slow. If the reheating is in or over a fire, the control is lacking 
and the piece must be quenched as soon as the proper tempera- 
ture is reached. This is usually determined by the color of the 
oxide or scale that forms on the brightened surface of the metal. 
The following table shows the temperatures, the corresponding 
colors, and the uses for which the different tempers are suited: 

430° Faint yellow 

450° Straw yellow Hardest and keenest cutting tools. 

470° Full yellow 

490° Brown yellow 1 Cutting tools requiring less hardness 
or orange j and more toughness. 

510° Purplish 1 Tools for softer materials, or those 

530° Purple J required to stand rough usage. 

Spring temper. Used for tools re- 
quiring great elasticity, and those 
for working very soft materials. 



550 Light blue 
560° Full blue 
600° Dark blue 



Special Steels. — In the common grades of steel the valuable 
properties are due to the presence of carbon, modified in some 
degree by other ingredients. There are other substances which, 
when united with iron in small proportions, give to the combina- 
tion increased strength, hardness, or other valuable properties. 
We have thus various special steels in which their properties 
may be due to the presence of carbon and other ingredients, or 
due chiefly to special ingredients other than carbon. The most 



ENGINEERING MATERIALS 1 99 

important of the special steels are known as nickel steel and 
tungsten steel. 

Nickel Steel. — An alloy known as nickel steel, containing 
about 3 per cent of nickel and varying amounts of carbon, is 
found to have increased strength and toughness as compared 
with ordinary steel. It is extensively used in the manufacture 
of guns and armor plate, and to some extent it has been em- 
ployed in government work for propeller shafts and for boiler 
plates. 

Tungsten Steel. — This steel, known also as Mushet steel, 
containing tungsten in proportions varying from 8 to 15 per 
cent, is very hard and can be forged only by the exercise of 
great care. Its hardness is not increased by tempering but is 
naturally acquired as the metal cools; hence it is said to be 
self-hardening. Some specimens contain also small amounts of 
manganese and silver. Its chief use is for lathe, planer, and 
other cutting and shearing tools where excessive hardness is 
required. 

Uses in Engineering. — Structural steel is used extensively in 
the construction of buildings and bridges and almost entirely 
in the construction of the hulls of modern ships. 

Cast steel, as well as cast iron, is used for pistons and cross- 
heads of engines, columns, bed plates, bearing pedestals and caps, 
propeller blades, and for many small pieces and fittings. 

Forged steel is used for columns, piston rods, connecting rods, 
crank and hne shafting, and for other parts of engines and 
machinery. 

84. Copper. — Copper in its pure state is red in color, soft, 
ductile, and malleable, with a melting point at about 2000 
degrees, and a tensile strength of from 20,000 to 30,000 pounds 
per square inch of section. It is not readily welded except 
electrically, but is easily joined by the operation of brazing. 
Attempts have been made to temper it, but without practical 



200 THE ELEMENTS OF MECHANICS OF MATERIALS 

success. It is readily forged and cast, and when cold, may be 
rolled into sheets or drawn into wire. When in sheets or in 
small pieces it may be spun, flanged, and worked under the 
hammer. 

The tensile strength of copper rapidly falls off as the tempera- 
ture rises above 400 degrees, so that from 800 to 900 degrees its 
strength is only about one-half that at ordinary temperatures. 
This pecuHarity of copper should be borne in mind when it is 
used in places where the temperature is liable to rise to these 
figures. If copper is raised nearly to its melting point in con- 
tact with air, it readily unites with oxygen and loses its strength 
in large degree, becoming, when cool, crumbly and brittle. Cop- 
per in this condition is said to have been burned. The possi- 
bility of thus injuring the tenacity of copper is of the highest 
importance in connection with the use of brazed joints in steam 
pipes. 

Copper unalloyed is used chiefly for pipes and fittings, espe- 
cially for junctions, elbows, bends, etc. For large sizes of pipes 
and fittings, the copper is made in sheets, bent and formed to 
the desired shape, and brazed at the seams; for small sizes the 
same general process is followed, or the metal is drawn from the 
solid and bent as desired after the drawing. 

Copper is also used as the chief ingredient of the various 
brasses and bronzes. 

85. Lead. — Lead is a very soft, dense metal, grayish in 
color after exposure to the air, but of a bright silvery luster 
when freshly cut. Commercial lead often contains small amounts 
of iron, copper, silver, and antimony, and when so combined is 
harder than the pure metal. It is very malleable and plastic. 
In engineering, lead is chiefly of value as an ingredient of bear- 
ing metals and other special alloys. Lead piping is also used to 
some extent as suction and delivery pipes for water where the 
pressure is only moderate, and where the readiness with which 



ENGINEERING MATERIALS 20I 

it may be bent and fitted adapts it for use in contracted 
places. 

86. Tin. — Tin is a soft, white, lustrous metal with great 
malleability. Commercial tin usually contains small portions 
of many other substances, such as lead, iron, copper, arsenic^ 
antimony, and bismuth. It is largely used as an alloy in the 
various bronzes and other special metals. Tin resists corrosion 
well, and in consequence is frequently used as a coating for con- 
denser tubes. It is also used for coating iron plates, the product 
being the so-called tin plate of commerce. It melts at about 
450 degrees, which corresponds to a steam pressure of 400 pounds 
per square inch, approximately. Due to this low melting point, 
tin is often used in the composition for safety plugs in boilers. 

87. Zinc. — Zinc, or '' spelter," as it is often commercially 
called, is a brittle and moderately hard metal with a very crys- 
talline fracture. The impurities most commonly found in zinc 
are iron, lead, and arsenic. It is used chiefly as an ingredient of 
the different brass and bronze alloys, and for coating iron and 
steel plates and rods. The process of applying zinc for such a 
coating is called galvanizing, and the product is '^ galvanized " 
iron or steel. Electricity is not used in the process; the articles, 
after being well cleaned, are simply dipped into a tank of melted 
zinc and then withdrawn. 

88. ALLOYS. — A mixture of two or more metals is called an 
alloy. The properties of an alloy are often surprisingly different 
from those of its ingredients. The melting point is sometimes 
lower than that of any of the ingredients, while the strength, 
elastic limit, and hardness are often higher than for any one of 
them. 

Mixtures of copper and zinc are called brass. Mixtures of 
copper and tin, or of copper, tin, and zinc, with sometimes other 
substances in small proportions, form gun metals, compositions y 
and bronzes. These terms are rather loosely employed. 



202 THE ELEMENTS OF MECHANICS OF MATERIALS 

Brass and compositions are used for piping and pipe fittings; 
globe, gate, check, and safety valves; condenser tubes and shells. 
The bronzes are employed for many of the uses of brass where 
more hardness, strength, and rigidity are required. They are 
used with success as a material for propeller blades. 

89. Timber. — Timber is not extensively used in modern en- 
gineering construction. The advance made in the production of 
steel, whereby its homogeneity is assured and its superior strength 
unquestioned, has caused it to be substituted for wood whenever 
it is possible and profitable to do so. 

Generally speaking, the heaviest and darkest colored timber 
is the strongest, and, in all cases, the strength of timber is great- 
est in the direction of its grain. 

The locality from which timber comes, the season of its cutting, 
and the duration of its seasoning, are factors in its strength, and 
the uncertainty arising therefrom makes it advisable to use a 
factor of safety of not less than 10 in calculations relating to the 
dimensions of timber to bear given loads. 

90. Concrete. — Concrete is a mechanical mixture of cement, 
sand, and broken stone or gravel, in the proportions, usually, of 
I : 2 : 4 or I : 3 : 6. It is largely used for laying foundations 
for buildings and bridges in wet ground, and for breakwaters and 
sea walls. After laying, it soon hardens to a strong mass which 
is little permeable to water. 

Concrete is not in the market as a manufactured product but 
must be made as needed. Whatever the proportions used, there 
is the utmost necessity for thorough mixing, water being added 
as may be necessary to secure coherency in the mixture. The 
use of machines designed for the purpose secures a more perfect 
mixture than that attained by the hand process. 

91. Reinforced Concrete. — Beams and columns of re- 
inforced concrete are products designed to avoid the uncer- 
tainty concerning the protection from corrosion and fire 



ENGINEERING MATERIALS 203 

that attends the use of the skeleton steel frame in building 
construction. 

There can be no question as to the appropriate use of steel for 
constructive purposes in such open structures as bridges and 
steamships, but in cases where a steel skeleton is vested with 
the strength of a structure, and is subsequently incased in 
terra cotta, stone, or brick, precluding visual inspection, there 
is serious question as to the propriety of its use. 

The use of concrete for constructive purposes was common 
among the ancients, and the fact that in some ruins, as they 
stand to-day, the concrete parts remain — the stone having 
long since disappeared — is conclusive evidence of its durabil- 
ity. There is abundant evidence also that iron embedded in 
concrete is protected from the corrosive influence of moisture 
and from the ravages of fire. 

It has been demonstrated experimentally that concrete is 
strong in compression but quite weak in tension. In the case 
of a concrete beam supported at the ends and loaded at the 
middle, it has been shown that the upper side, which is in 
compression, is capable of supporting ten times the load 
which would cause failure at the lower side, which is in 
tension. 

Having in steel a material of very high tensile strength, and 
in concrete a material possessing high compressive strength as 
well as the properties of durability and impermeability to 
moisture, the problem arose of effecting their combination so 
as to produce a composite material having the desirable quaHties 
of both. The solution of this problem was the production of 
reinforced concrete. 

In the case of beams, steel bars are embedded in the area of 
the concrete below the neutral axis to reinforce the concrete 
subjected to tensile stress, thus enabHng the full strength of the 
compression area above the neutral axis to be utilized in sup- 



204 THE ELEMENTS OF MECHANICS OF MATERIALS 

porting heavier loads than would have been possible without the 
combination. 

In the case of columns, the well-known fact that the most eco- 
nomical metal section is that of the hollow cylinder suggested 
at once the conception of a column having three concentric 
parts, viz., a central core of concrete, an intermediate zone of 
steel, and an outer zone of concrete. The steel is thus protected 
from fire and moisture and is best disposed for the utilization of 
its maximum strength. 

Between the tubular reinforced concrete column just described 
and the plain non-reinforced concrete column there are a variety 
of possible combinations. In the form generally used, appHcable 
ahke to columns and piles, the reinforcement consists of vertical 
rods of steel tied together by a system of horizontal wires. 
These horizontal ties not only keep the vertical rods in position, 
but materially assist in preventing flexure in them; they also 
prevent lateral bulging of the concrete. 

The reinforcement for both beams and columns is first placed 
in position, after which they are enveloped by a wooden form, 
or mold, into which the concrete is dumped, and rammed when 
necessary. After a period of thirty hours the form may be 
removed and the product allowed to season for several weeks 
before being subjected to its load. 

92. Adhesion of Concrete to Steel. — It has been shown by 
experiment that the concrete on the compression side of rein- 
forced concrete beams may be subjected without rupture to a 
stress twenty times that which would cause failure in a tension 
test in the concrete alone. It has also been shown that rein- 
forced concrete acquires a power to resist crushing which is 
greater than the sum of the resistances of the two materials 
taken separately. Such remarkable results could not be obtained 
were it not for the adhesion between the concrete and steel, such 
adhesion offering resistance to sliding between the two surfaces 



ENGINEERING MATERIALS 205 

and facilitating the transference of the forces from one surface 
to the other. 

The bond between the steel and concrete occasioned by ad- 
hesion alone is liable to be destroyed by internal stresses due to 
shocks and vibrations, and from unequal expansions resulting 
from thermal changes, and it is with the idea of strengthen- 
ing the adhesion bond by mechanical means that the rein- 
forcing bars are usually twisted or have projections on their 
surfaces. 

93. Proportion of Reinforcement. — To secure a uniform dis- 
tribution of the stress in a reinforced concrete section, the rein- 
forcement should consist of steel bars of small section distributed 
so that each shall bear its allotted part of the stress. The pro- 
portion of steel to concrete depends directly upon the ratio of 
the coefficients of elasticity of the two materials. The value of 
E for steel is 30,000,000 pounds per square inch, implying that 
a force of one pound would extend or compress a bar of steel 

I square inch in area by of its original length. The 

30,000,000 

value of E for concrete may be taken as 3,000,000 pounds per 
square inch, so that for an equal extension of the two materials 
the steel will bear ten tim§s the stress that can be borne by the 
concrete. 

Suppose a bar of steel i square inch in section area to be sur- 
rounded by a ring of concrete i square inch in area; and suppose 
further that the steel be subjected to a direct pull that would 
occasion in the concrete the safe allowable tension stress of 50 
pounds per square inch. The elongation in the concrete would 

then be 50 X = 0.0000167 inch. Considering the bond 

3,000,000 

between the steel and concrete to be perfect, the. steel would 

suffer an equal elongation, occasioning a stress in the steel of 

0.0000167 X 30,000,000 = 500 pounds per square inch, 



2o6 THE ELEMENTS OF MECHANICS OF MATERIALS 

a result only one-thirtieth of the safe allowable tension stress 
for steel; consequently, the sectional area of the steel may be 
reduced to one-thirtieth square inch, thus raising its stress to 
15,000 pounds per square inch without causing failure in the 
xconcrete. 



CHAPTER X 
TESTING MATERIALS 

94. Stress. — The application of external forces to a piece of 
material tends to change its shape, and this tendency induces 
internal forces, known as stresses, which offer resistance to the 
change. These stresses may be of three kinds: 

1. If the external force be applied at right angles to the 
section, and acts away from it, the stress is one of tension, or a 
tensile stress. 

2. If the external force acts toward the section, the stress is 
one of compression, or a compressive stress. 

3. If the external force acts parallel to the section, the stress 
is one of shear, or a shearing stress. 

It is a fundamental assumption that these direct stresses are 

uniformly distributed over the section, so that if W denotes the 

external force or load, A the area of section, and 5* the unit 

stress, we must have, in the absence of rupture, W = AS. W is 

usually expressed in pounds and A in square inches, so that we 

W 
shall have for the unit stress 5 = — in pounds per square inch. 

95. Strain. — A piece of material which is stressed by the 
application of external force undergoes some change in its dimen- 
sions, either lengthened or shortened, and the amount of this 
distortion is known as the strain due to the external force, or 
load. 

96. Different Kinds of Tests. — Materials are tested for ten- 
sion by puUing apart a test piece of specified dimensions; for 
compression, by crushing a piece of definite dimensions; for 

207 



208 THE ELEMENTS OF MECHANICS OF MATERIALS 

transverse strength, by supporting a piece at two points and 
breaking or bending it in a testing machine by applying a load 
at an intermediate point; for torsion, by twisting apart a piece in 
a machine designed for the purpose ; for direct shearing, by break- 
ing a riveted or pin-joint connection in a machine; for impact, 
or shock, by letting a weight drop through a definite height, and, 
by its blow, develop suddenly the stress in the material. 

97. Ultimate Strength. — The ultimate strength of a test 
piece is the load required to produce fracture, reduced to a 
square inch of original section; or, in other words, it is the ulti- 
mate or highest load divided by the original area. Thus, if the 
area of the cross section of a test piece is 0.42 square inch, and 
the load producing fracture is 28,400 pounds, the ultimate 

strength is — — — = 67,620 pounds per square inch. 
0.42 

98. Elastic Limit. — The elastic limit is the load per square 
inch of area that will just produce a permanent set in the mate- 
rial. Thus, in a tension test, if the cross-section area of the test 
piece be 0.7 square inch, and a permanent set just be produced 
by a load of 28,000 pounds, the elastic limit is 40,000 pounds 
per square inch. 

99. Factor of Safety. — - To insure safety in engineering con- 
struction the stresses due to the working load must not exceed 
the elastic limit of the material used, and to insure this provi- 
sion it is customary to make the working load very considerably 
less than the load necessary to produce fracture. The ratio 
between the breaking load and the safe allowable load is the 
factor of safety; or it is the quotient obtained by dividing the 
ultimate strength by the working stress. 

100. Elongation. — The increase in length of a test piece, 
measured just before rupture, divided by the original length of 
the piece, expresses the elongation. 

When a load is first applied to a test piece the elongation 



TESTING MATERIALS 209 

is nearly uniformly distributed throughout the whole length. 
This continues until the piece begins to contract in area near 
the point of final rupture, and nearly all the subsequent elon- 
gation is restricted to the immediate vicinity of this point. 

In expressing the elongation of any material, the length of 
the test piece must be stated. If the length of the test piece 
is 8 inches, and an extension in length of 2 inches is noted just 
before rupture, the elongation is then expressed as 25 per cent 
in 8 inches. 

loi. Reduction in Area. — The reduction in area is found by 
dividing the difference between the original and final section 
areas at the point of rupture by the original area, expressing the 
fraction in per cent. 

102. Testing Machines. — Machines devised to determine the 
physical properties of materials are of two general classes, — 
hydraulic and screw gear. With either class the load is applied 
to the specimen to be tested in a manner that enables it to be 
read instantly on some form of weighing machine. 

The weighing system employed with the hydraulic machine 
consists usually of registering the appHed load by means of a 
gage which records the pressure within the cylinder of an 
hydraulic press. Hydraulic machines, with the exception of 
the one devised by Emery, are lacking in sensitiveness and ac- 
curacy and are not extensively used. 

Of the screw gear type of machines, those of Riehle and 
Olsen are very generally used. With either machine the loads 
are appHed from some external source and are transmitted by 
means of spur and bevel gears to upright screws. 

The Riehle machine consists of two heads, one fixed and the 
other movable. The upper head is fixed and is supported by 
two cast-iron columns which rest on the weighing table of 
the machine. The weighing table rests on steel knife-edges in 
the levers of a compound system, the last lever of which is the 



2IO THE ELEMENTS OF MECHANICS OF MATERIALS 

weighing beam of the machine. Two upright pulling screws, 
which turn in long bearings and have the main gears keyed to 
their lower ends, pass through nuts in the movable lower head 
and reach nearly to the under side of the upper head. The 
screws raise or lower the movable head according to the direc- 
tion in which they revolve. 

In making a tensile test, the specimen is gripped at its end 
by jaws in the two heads. Power is applied from an external 
source, either hand or motor, in a direction to lower the movable 
head by the screws, thus transmitting a pull to the specimen, 
thence to the upper head, and then to the weighing table. The 
pressure on the weighing table is transmitted through the lever 
system to the weighing beam. On the weighing beam is a 
counterpoise which the operator moves along the beam to main- 
tain the balance as the load is gradually applied, and thus has 
constantly under observation the magnitude of the applied load. 

In making a compression test, a cyhndrical cast iron block is 
bolted to the imder side of the movable head and the specimen 
is placed between that and a similar block on the weighing table. 
As the screws lower the movable head the pressure is brought 
to bear on the specimen, thence through the weighing table and 
lever system to the weighing beam. 

In making a transverse or bending test, the specimen is laid 
on two supports which rest on the weighing table. As the 
movable head is drawn doTVTi by the screws, a projection on 
the under side of the head bears on the middle of the specimen 
and the pressure is transmitted to the weighing beam, as in the 
other tests. The deflections of the specimen may be measured 
to o.ooi of an inch by attaching an instrument known as a 
deflectometer, or transverse indicator. 

103. Forms of Test Specimens. — Test specimens are made 
from coupons cut from the finished product of the material to 
be tested, and are of prescribed form. The American Society 



TESTING ]MATERIALS 



211 



for Testing Materials recommends the form of specimen shown 
in Fig. 112 for tensile tests of metal plates, inch lengths being 



,A 



Parallel section 



not less than 9" 



J^*-i->r-i— 



-about-l&^ 



Fig. 112. 

marked with center punch on the length of parallel section. 
Figure 113 shows the form prescribed by the Navy Department. 



about 18- 



Fig. 113. 

The round form of test piece shown in Fig. 114 is that used 
for testing manufactured products other than plates, such as 
shafts, axles, and beams. 



3 © C 



-about 18- 



Fig. 114. 



Figures 115 and 116 illustrate the cold bending and angle 
tests for wrought iron and steel. The material must stand 
these tests without sign of fracture. 



212 THE ELEMENTS OF MECHANICS OF MATERIALS 

The tensile test of metals is the simplest and most important, 
as it determines the physical properties of ultimate strength, 
yield point, elastic Kmit, and percentage of elongation. 



en- — , 





Fig. 115. Fig. 116. 

104. Ductility. — Materials such as wrought iron, mild steel, 
copper, and other metals which may be lengthened by the appH- 
cation of an external force, with a corresponding decrease in 
thickness or in diameter, possess the property of ductihty. 

105. Plasticity. — If in the process of stretching a ductile 
material the load be removed and none of the strain disappears, 
the material is said to have passed beyond the ductile and to 
have entered the plastic stage. The plasticity of a material is 
determined by the final elongation and contraction in area of 
the test piece. In structures subjected to hve loads and shocks 
it is as important to know the power of the material used to 
resist deformation as it is to know its ultimate strength, and 
for that reason specifications for iron and steel usually require 
a certain percentage of elongation and contraction of area in a 
stated length of test piece. 

106. Stress-strain Diagram. — If in testing a material the 
gradually applied loads be plotted as ordinates and the corre- 
sponding strains as abscissas, the resulting curve is known as a 
stress-strain diagram. 

For tension tests of wrought iron and mild steel such a dia- 
gram will take the form shown in Fig. 117. 

The load being gradually increased, it will be found that 
within a certain limit the strains, or extensions, will be directly 



TESTING MATERIALS 



213 



proportional to the augmentations in the load, and that if the 
stress be relieved the test piece will return to its original length. 
During this period the material is said to be perfectly elastic, and 
will be so represented in the diagram by the straight Hue OA. 
By continuing the gradual increase in the load a point will be 
reached where the proportionality between the strain and the 
augmentations in the load ceases, the strain increasing much 
more rapidly than the load; and if the stress be reHeved the 



Max. stress 



Breaking 

stress 




piece will not return to its original length, but will acquire a 
permanent set. The load at which this occurs is known as the 
elastic limit of the material. 

A further increase in the load very soon develops a point 
where the extension increases very rapidly, — as much as from 
10 to 15 times its previous amount, — known as the yield point. 
This rapid increase in the extension usually occasions an appar- 
ent reduction in the stress, as shown by the fall in curvature 
beyond the point B. For commercial purposes the yield point 
and the elastic Hmit are taken as the same point, and the ordi- 



214 THE ELEMENTS OF MECHANICS OF MATERIALS 

nate at B would represent, to the scale of the diagram, this limit 
in pounds per square inch of section of the material. 

Passing the yield point, the strains increase much faster than 
the loads, but if the stress in the material be reheved a careful 
measurement will show the disappearance of a small portion of 
the extension, indicating the existence still of some elasticity 
and that the specimen is passing through the ductile stage. 

At about the time the maximum stress is reached at C, the 
material appears to have reached the plastic state, the extension 
increasing, in time, without increase in load. Up to this point 
the strain has been evenly distributed throughout the length of 
the specimen, but here occurs an extension and reduction in 
section purely local, immediately followed by rupture at D. 

Within the elastic limit the extension of the specimen probably 
would not exceed o.ooi of its length, so it is quite impossible 
to make direct measurements of the extensions corresponding to 
the augmentations in load. Some form of extensometer is used 
to make these measurements. 

107. Extensometer. — The instrument designed to measure 
accurately the minute extensions within the elastic Hmit of a 
specimen during a tensile test is known as an extensometer. 
There are various forms of this instrument, but the type known 
as the Richie- Yale is in very general use and gives dependable 
results. 

It consists essentially of two clamps, which are fastened to 
the test specimen by set screws. The paralleHsm of the clamps 
is secured by a squaring gage bar which fits neatly in guides in 
the clamps, a set screw arrangement permitting the distance 
between the clamps to be varied according to the distance be- 
tween the punch marks on the specimen — usually the standard 
distance of 8 inches — in order that the set screws of the clamps 
may fit exactly in the punch marks. The upper clamp has two 
projecting arms, 180 degrees apart, through which pass two in- 



TESTING MATERIALS 21 5 

sulated bars which are connected with the lower clamp in circuit 
with a battery and bell. The lower clamp has projecting arms 
corresponding with those of the upper clamp, and through them 
pass two micrometer screws, each having a vertical fleet of one 
inch and reading to o.oooi of an inch. 

The instrument being attached to the specimen as indicated, 
and the squaring gage bar removed, it is ready for use. An 
increment of load being applied, the elongation of the specimen 
is measured by taking the reading, first of one of the micrometer 
screws and then of the other, by running them up until the con- 
tact of the point of the screw with the insulated bar completes 
the circuit and causes the bell to ring. Another increment of 
load is then added and the readings taken again, and so on until 
the elastic limit is reached, the average readings of the two 
micrometer screws for each increment of load giving the actual 
elongation of the specimen. The instrument is removed from 
the specimen after the elastic limit is passed. 

108. Tensile Test of Steel. — Figure 118 is an illustration of 
a tensile test of steel, the scales being small in order to keep the 
diagram within the limits of the page. 

The original dimensions of the test piece were: Length, 
8 inches; diameter, 0.75 inch; area of section, 0.4418 square 
inch. 

The final dimensions were: Length, 10.2 inches; diameter 
0.4843 inch; area of section, 0.1842 square inch. 

An inspection of the diagram shows the elastic Hmit to have 
been reached at about 21,000 pounds. 

Elastic limit of specimen = — - — - = 47,530 pounds per square 

0.4418 

inch. 

TTi.- ^ ^ Maximum load 20,700 . , 

Ultimate stress = ^ . . — = -^^ — = 67,220 pounds 

Original area 0.4418 

per square inch. 



2l6 THE ELEMENTS OF MECHANICS OF MATERIALS 

T7 ^ • • O • -U (10.2 — 8) 100 

Extension in 8 inches = ^ — = 27.5 per cent. 

8 

r- 4. 4.- r (0441 8 - .1842) 100 _ 

Contraction of area = ^^^—^ -=^-^ = 58.31 per cent. 

0.4418 



30 000 
28 
26 
24 
^3 
20 
18 
16 
14 
12 
10 



::: 


rmTMM 
5 

:±z : : 


■^-A^ L 
^ L_ 

::|:ffi:::T 

:::::Eee;|e± 


Ni|' hi 

::::::±:::: 
: i_: : ::n: 


:i:::::::5 

:::::: + ::: 

(-- 

t:::::::::: 

iT-±E — 

WW 


m 

\z:::::: 

m 

::::::::: 



ScalesiXoads, 0.2=2000 Ibs.j ExteBsions, full size- 
Fig. ii8. 



Loads. 


Extensions. 


Remarks. 


2,000 


0.0012 




4,000 


0.0024 




6,000 


0.0034 




8,000 


. 0047 




10,000 


. 0060 




12,000 


0.0072 




14,000 


. 0083 




16,000 


0.009s 




18,000 


0.0106 




20,000 


O.OI18 




22,000 


0.0220 


Yield point 


26,000 


0.5300 




29,680 


1.9200 


Maximum stress 


25,820 


2 . 2000 


Breaking stress 



TESTING MATERIALS 



217 



The load was gradually applied with the uniform augmenta- 
tion of 2000 pounds, and the data of the test was tabulated as 
shown in the table. 

To find the modulus of elasticity we proceed as follows : 
The sum of the extensions up to and including the 18,000 
pound load — a point well within the Kmit of elasticity — is 
0.0533 inch, and the sum of the loads is 90,000 pounds. The 

mean extension for 2000 pounds is, therefore, ^"'^ = 0.0012 

45 
inch. 



By Art. 31, p. 63, we have E 



SL 



in which L is the original 



length and S the load producing the extension y. Here L = 8 
inches, 5 = 2000 pounds, and y = 0.0012 inch. 

8 X 2000 



Hence, E = 



= 30,180,000 lbs. per sq. inch. 



0.4418 X 0.0012 

109. Compression Tests. — In testing materials for compres- 
sion, the specimens are not longer than 1.5 to 3 times the diam- 
eter. If the specimens are long the failure under compression 
will be by buckling or bending, and for intermediate lengths 
partly by crushing and partly by bending. 

AVERAGE PHYSICAL PROPERTIES OF MATERIALS. 





Pounds per square inch. 


Pounds. 


Material. 


Elastic limit. 


Ultimate strength. 


Modulus 
of elas- 
ticity. 


Weight 




Ten- 
sion. 


Com- 
pres- 
sion. 


Shear- 
ing. 


Ten- 
sion. 


Com- 
pres- 
sion. 


Shear- 
ing. 


cubic 
foot. 


Hard steel . . 


60,000 

36,000 

25,000 

90,000 

170,000 

28,000 

6,000 

7,000 






100,000 
60,000 
50,000 
120,000 
170,000 
50,000 
25,000 
30,000 
300 
10,000 






31,000,000 
30,000,000 
27,000,000 
31,000,000 
33,000,000 
25,000,000 
15,000,000 
15,000,000 
3,000,000 
1,500,000 


490 
490 
490 


Mild steel 

Cast steel 


33,000 


24,000 


60,000 


50,000 


Tool steel, unhardened 










490 


Tool steel, hardened. . . 










490 


Wrought iron 


20,000 
24,000 
1,000 


25,000 


55,000 
90,000 
49,000 
2,500 
8,000 


40,000 
20,000 

1,400 
600 


480 


Cast iron 


450 




550 


Concrete 


150 


Timber 


3,000 


40 











PART II 

THE ELEMENTS OF POWER 
TRANSMISSION 



CHAPTER I 
TRANSMISSION OF POWER BY BELTS AND ROPES 

1. Flat Belt Gearing. — The transmission of power by means 
of belts running over pulleys is an important and familiar me- 
chanical contrivance. It is practically noiseless, and may be 
used for transmitting power through a distance as great as 
30 feet without intervening support. For greater distances 
idle or binder pulleys are generally used to tighten the slack 
side of the belt. 

The principal disadvantage of belt drives is that due to the 
slip occasioned by the freedom of the belt to slip over the pulley, 
rendering the transmission not so positive as that through the 
medium of gear wheels; but this disadvantage becomes an 
advantage in preventing shocks in cases where mechanisms 
at rest are suddenly thrown into gear. 

2. Materials for Flat Belts. — The most common material 
for flat belting is leather, though cotton and India rubber are 
not infrequently used. The best leather belting is made of 
oxhide, the strips of the hide being tapered at the ends and 
cemented together under great pressure, and then laced or 
riveted with copper rivets and washers. The thickness of the 
single-ply belt varies from yig- to A inch. If a greater thick- 

219 



220 THE ELEMENTS OF POWER TRANSMISSION 

ness is required, two or three strips are cemented together 
under pressure, thus producing two-ply and three-ply belting 
in varying thickness up to i inch. The average weight of 
leather belting is 0.036 pound to the cubic inch. 

Cotton belts are usually made by stitching together canvas 
or ducking, but are sometimes woven solid. Such belts are 
cheaper and equally as strong as those made of leather, but the 
coefficient of friction is rather low unless the material is properly 
sized. The weight of cotton belting varies with the material, 
but the average of a cubic inch may be taken as 0.034 pound. 

Rubber belting is produced by treating canvas with a com- 
position of rubber in such manner as to fill all its interstices. It 
is then wrapped in rubber and vulcanized under heat and pres- 
sure. It is adaptable to use in damp places, since the material 
is not affected by moisture. 

The weight of rubber belting is about 0.044 pound per cubic 
inch. 

3. Strength of Leather Belting. — The ultimate strength of 
leather used for belting varies from 3600 to 5000 pounds per 
square inch of section and the strength of the laced joint may 
be taken as one-third that of the solid leather. Taking a factor 
of safety of 5, the safe working stress of a laced joint varies from 
240 to 330 pounds per square inch of section for single-ply belting, 
and from 500 to 600 pounds per square inch of section for double- 
ply. The width of the belt must be made sufficient to with- 
stand the tension. The safe tension in pounds per inch of width 
of single-ply belting ranges from 50 to 80 according to the 
thickness and the safe stress of section of the material. 

4. Coefficient of Friction of Leather Belting. — The coefficient 
of friction of leather belting running over smooth iron pulleys 
is a very important but exceedingly variable quantity, ranging 
from 0.22 to 0.35. With wooden pulleys having varnished 
faces the coefficient of friction is somewhat greater. 



TRANSMISSION BY FLAT BELTS AND ROPES 221 

5. Velocity of Leather Belting. — The velocity of leather 
belting varies from 2000 to 6000 feet per minute, but the most 
economical speed is from 4000 to 5000 feet. At higher speeds 
the effect of centrifugal action is excessive and the life of the 
belt shortened. 

6. Velocity Ratio in Flat-Belt Transmission. — If there is no 
slipping of the belt on the pulleys when transmitting motion 
from one pulley to another, the outer surface of the rim of each 
pulley will have the same velocity as the belt; so if we denote 
by Di and D2 the diameters of the driver and follower pul- 
leys respectively, and by Ni and N2 their revolutions per minute, 
we shall have 

Speed of outer surface of driver rim = ttA ^^i, 
and Speed of outer surface of follower rim = TD2N2, 

and since each of these is equal to the speed of the belt, we have 
ttDiNi = irD^N^, whence ~ = ^' 

That is, the velocities of the pulleys are inversely as their 
diameters. 




Fig. I. 

When motion is transmitted from one shaft to another by 
belting, one or more shafts intervening, as in Fig. i, we shall have 

N2 Di N^ A' ^"^ N^ A* 



2 22 THE ELEMENTS OF POWER TRANSMISSION 

Taking the product of these equations, member by member, 
and remembering that N2 = N3, and N^ = N5, we have 

iVi ^ A A A 

N, A A A* 

That is, the ratio of the speed of the first pulley to the speed 
of the last pulley is equal to the continued product of the ratios 
of the diameters of each pair of pulleys taken in order. 

The results just obtained are true only for very thin belts, 
or for belts whose thickness is so small in comparison with the 
diameters of the pulleys as to be negligible. The thickness of 
belts ordinarily used may have an appreciable effect on the 
velocity ratio of two pulleys. While the belt is in contact with 
the pulley, its inner surface is in compression and its outer 
surface in tension, but the neutral surface midway between is 
of constant length. It follows that the velocity of the surface 
of the belt in contact with the pulley is less than the velocity of 
the neutral surface, and that the true velocity ratio of two 
pulleys is obtained by increasing the diameters of the pulleys 
by an amount equal to the thickness of the belt. 

7. Slip of Flat Belts. — Another error in the velocity ratio 
of belt-driven pulleys is due to slip. The driving side of a belt 
is necessarily stretched more than the slack side, and in con- 
sequence the driving pulley receives a greater length of belt 
than it gives off to the follower pulley; therefore the speed of 
the driving side is a trifle greater than that of the slack side. 
But the speed of the rim of a pulley is the same as that of the 
belt it receives; therefore the speed of the rim of the driver will 
be somewhat greater than that of the rim of the follower, and, 
as a consequence, the speed of the follower will be less than that 

given by the ratio equation ~ = -~- This difference between 

the speeds of the rims of the driver and the follower is known 
as slip, and amounts to about 2 per cent. 



TRANSMISSION BY FLAT BELTS AND ROPES 223 

Example I. — The pulley on the shaft of a steam engine is 
4 feet in diameter, from which a belt passes to a pulley 2 feet 
in diameter on a shaft in a room above; a belt passes from an- 
other 2-foot pulley on this shaft to one of 10 inches in diameter 
on a third shaft; an 18-inch pulley on the third shaft is belted 
to a 6-inch pulley on the spindle of a dynamo. Find the speed 
of the dynamo when the engine is making 160 revolutions per 
minute. 

Solution. - Speed of engine ^24^10^^ = ^, 
bpeed of dynamo 48 24 18 72 
whence 

Speed of dynamo = — = 2304 revolutions per minute. 

If the thickness of the belts in this example were 0.25 inch and 
the slip 2 per cent, we should have 

Speed of dynamo = ^ ' ^ X "^ X -— ^ X 160 X 0.08 = 2155 
24.25 10.25 6.25 

revolutions per minute, showing an error of 6.9 per cent in com- 
puting the speed of the dynamo by the usual method. 

When great accuracy is required, the errors due to sKp and 
to the failure to include the thickness of the belt in the pulley 
diameters should be corrected. 

8. Tensions in Flat Belts. — When a belt is fitted to two 
pulleys it is strained over them while at rest with a tension Tqj 
which is uniform throughout the belt. When motion ensues, 
the driving side of the belt stretches and its tension Ti increases. 
At the same time the slack side of the belt is shortened and its 
tension T2 decreases. The theory of belting rests on the assump- 
tion of perfect elasticity in the belts, so that the lengthening 
of the driving side of the belt must equal in amount the shorten- 
ing of the slack side, and the average tension in the belt remains 
constant. That is, we should have 

T, + T2 = 2 To. 



224 THE ELEMENTS OF POWER TRANSMISSION 

Such is not the case, however, as the materials used for belting 
are far from being perfectly elastic; but the error involved is 
not of such consequence as to destroy the theory when applied 
to the conditions under which belts are ordinarily used. 

9. Frictional Resistance between a Flat Belt and a Pulley. — 
Let the two pulleys of Fig. 2 be joined with an endless belt. 
If a force tends to turn the driving pulley D in the direction 
indicated by the arrow, the lower part of the belt will be stretched 




Fig. 2. 



SO that its tension Ti will be increased, and the tension T2 in 
the upper part of the belt will be decreased an equal amount. 
Each element of the belt in contact with the pulley assists the 
action of the tension in the slack side of the belt in resisting the 
tension in the tight side, and therefore the tension in that part 
of the belt in contact with the pulley varies at every point. 
When the difference, Ti — T2, becomes sufficient to overcome 
the resistance to motion in the driven pulley F, its rotation 
begins. The difference in tensions in the two sides of the belt 
is the amount of friction between the belt and the pulley, and is 
the measure of the driving force. If the tensions were equal, 
their moments about the center of the driven pulley would be 
equal and there would be no rotation, since there would be equal 
turning tendencies in opposite directions. 

Let Ti and T^., Fig. 3, denote respectively the tensions in the 
tight and slack parts of the belt that are not in contact with 



TRANSMISSION BY FLAT BELTS AND ROPES 



225 



the pulley. Suppose Jo: to be a very small part of the central 
angle a subtended by the arc of contact of the belt. 




Fig- 3- 



Considering the equiHbrium of the very small arc subtending 
Ja, we will assume the tensions at its extremities to be T and 
T -\- dT. If the resulting reaction between the belt and the 
pulley rim due to these tensions be denoted by R, we shall have 
the static equation 

R = 2 T sin — = Tday 

2 

since da is very small and sin — and — are very approximately 

the same. 

As slipping is about to take place, ^r is the measure of the 
friction over the small arc considered. 

Then dT = y.R = fxTda, 

in which ^ is the coefficient of friction. 



226 THE ELEMENTS OF POWER TRANSMISSION 

■jrp 

For the small arc considered, ^ = ixda, and for the whole 
arc of contact we have 



UTo J- *Jo 



T2 
from which 

T 

log Ti - log T2 = f^a, or log -7 = Aia, 

T 

whence -7 = e'^, 

T2 

in which ^ = 2.718, the base of the Naperian system of logarithms, 
and a is expressed in circular measure. If a is given in degrees 

it may be converted into radians by multiplying by — — • 

180 

Hence, it is seen that the ratio of the tensions depends only 
upon the coefhcient of friction and the angle at the center sub- 
tended by the arc of contact of the belt, and is independent of 
the diameter of the pulley. 

A convenient expression for the ratio of the tensions is 

T 

logioT^ = Mo: logio 2.718 = 0.4343 fxa. 
^2 

The result just obtained is applicable to a rope making several 
turns about a post, and explains how it is possible for one man 
to check the headway of a ship when docking by taking several 
turns of a hawser about a post on the dock and pulHng the slack 
end with a comparatively small force. 

Example II. — A hawser from a ship makes three turns about 
a post and is pulled with a force of 50 pounds. The coefficient 
of friction being 0.35, what force is exerted in checking the 
headway of the ship? 

Solution. — 

Here r2 = 50, a = 2 tt radians, and /z = 0.35. 



TRANSMISSION BY FLAT BELTS AND ROPES 227 

Then log|-' = 0.4343 X 0.35 X 67r = 2.86638, 

^ 2 

whence 

T 

-7^ = 735-15, and Ti = 735.15 X 50 = 36,757 pounds. 

10. Transmission of Power by Flat Belts. — In a belt con- 
nection between two pulleys the arc of contact to be considered 
is that of the smaller pulley, and if the connection is by open 
belt the arc of contact of the smaller pulley will be less than 
180 degrees, but with a crossed belt the arc of contact will be 
greater than 180 degrees. It follows that with the same con- 
ditions, a greater power may be transmitted with the crossed 
belt. The crossed belt, however, is subject to great wear, and 
generally its use is restricted to cases where it is desired to have 
the pulleys turn in opposite directions. 

The friction between a belt and pulley limits the power that 
can be transmitted. When overloaded, a belt will slip rather 
than break; therefore at the point of slipping the driving force 
is Ti — T2, and we shall have 

H.P. transmitted = (Jl^l^ = (liJzMll , 
33,000 550 

according as to whether the velocity is expressed in feet per 
minute or in feet per second. 

The ratio of the tensions is given by 

T 

log -r = 0.4343 fxa. 

Example III. — An open belt 3 inches wide connects a pulley 
5 feet in diameter with one 18 inches in diameter, the distance 
between the pulley centers being 12 feet. The larger pulley 
makes 200 revolutions per minute, the coefScient of friction is 
0.35, and the greatest tension in the belt must not exceed 80 
pounds per inch of width. Find the horse power transmitted. 



228 THE ELEMENTS OF POWER TRANSMISSION 

Solution. — 

Velocity of belt in feet per second = 5 ^ X 200 ^ ^2.36. 



60 



Ti = 80 X 3 = 240 pounds. 




Fig. 4. 



Referring to Fig. 4, 



sin 6 



_ 30 - 9 
144 



0.1458, whence = 8° 23'. 



Arc of contact of smaller pulley = [180° —2 (8° 23')]-^ 

180 



= 2.85 radians. 



Ti 



log— = 0.4343MQ: = 0.4343 X 0.35 X 2.85 = 0.43320, 

^2 



therefore 



240 



Then 



rp rp 

-~ = 2.71, whence T2 = — - = ^^^^^ = 88.56 pounds. 
12 2.71 2.71 

HP _ (Ti -T,)v _ (240 - 88.56) 52.36 _ 
550 550 



II. Centrifugal Action in Belts. — When belts are run at 
high speed the tensions are greater than those due to the power 
transmitted on account of the centrifugal action in that part 
of the belt in contact with the pulley. This centrifugal action 
also diminishes the normal pressure of the belt on the pulley 
rim, and therefore decreases the frictional resistance. For 



TRANSMISSION BY FLAT BELTS AND ROPES 



229 



speeds beyond 3500 feet ' per minute the stress due to cen- 
trifugal action increases rapidly and should be taken into 
account. 

Let T denote the tension due to centrifugal action in the belt 
of Fig. 5, the arc of contact being a. 



dFsin(90-f+^) 




<7Fcos(9O-f+0) 



^Tsin(90-f) 



T cos (90- — ) 



Fig. 5. 



Consider the very small arc of the pulley subtending the angle 
dS, and let dF denote the centrifugal force set up in the part 
of the belt in contact with this arc. 

We have the static equation 

2 T cos (90° - - ) = r^^ sin (90° - - + A 
2rsin-= CdF cos(- -e\ 



or 



dF 



dmv^ 



.23<3 THE ELEMENTS OF POWER TRANSMISSION 

If Wi denotes the weight of the belt in pounds per linear foot, 

then dm = — - » rdd, 



and ^j,^W,rdet^W,^^ 



Then 



2 T sm - = / cos f d]de = -^^\ - sin [ d\\ 

WiV^ . a 

= • 2 sm - ' 

g 2 

, ^ 2 Wiv' 2rWi v^ Wv^ 

whence 2 i = = • - = ? 

g g r gr 

and T = > 

2gr 

in which W = 2rWi is the weight of a portion of the belt whose 
length equals the diameter of the pulley; v the velocity in feet 
per second; g the acceleration of gravity, 32.2 feet per second 
per second ; and r the radius of the pulley in feet. 

Example IV. — Find the width and thickness of belt neces- 
sary to transmit 10 horse power to a 15-inch pulley so that the 
greatest tension may not exceed 60 pounds per inch width of 
belt when the pulley makes 1200 revolutions per minute, the 
weight of the belt per square foot being 1.5 pounds, the coefficient 
of friction 0.25, and the arc of contact of the belt 165 degrees. 
The weight of a cubic inch of leather is 0.036 pound. 



Solution: 



TT X 15 X 1200 o £ 4. 1 

= — ; = 78. £54 feet per second. 

12 X 60 ' ^^ ^ 

H.P. per second = HunlAl, 



TRANSMISSION BY FLAT BELTS AND ROPES 23 1 

whence Ti — T2 = ^^-z = 70 pounds. 

78.54 

a = arc of contact = — ^- — - = 2.88 radians. 

180 

T 

We have, log-r = 0.4343 fxa = 0.4343 X 0.25 X 2.88 = 0.31270, 
i 2 

T 

whence -7 = 2.05. 

Then 2.05 Ti — T^ = 70, whence T2 = 66.67 pounds, 
and Ti = 2.05 X 66.67 = 136.67 pounds. 

The effect of centrifugal action increases the tension by 

2gr 

pounds. If w denotes the width of the belt in inches, then 

W = weight of a portion of the belt whose length equals the 

2 TW 

diameter of the pulley = 2 rWi = weight of square feet of 

12 

the belting = ^ pounds. 

12 

Then = — ^^^^^ = 23.05 w^, 

2 gr 2 gr X 12 

and Ti = 136.67 + 23.95 ^• 

But from the conditions of the problem the tension must 
not exceed 60 z£;; 

hence 6oze^ = 136.67 + 23.95 ze;, whence w = 3.79 inches. 

If / denotes the thickness of the belt in inches, then 144 t is 
the volume in cubic inches of i square foot of the belting, and 
since i cubic inch of leather weighs 0.036 pound, we have 

144 / X 0.036 = 1.5, whence t = 0.29 inch. 

12. Flat Belt Connections between Non-parallel Shafts. — 

Two non-parallel and non-intersecting shafts may be connected 
by a belt, provided the pulleys are so placed that the point at 
which the belt leaves either pulley lies in the plane of the other 



232 



THE ELEMENTS OF POWER TRANSMISSION 



pulley, and provided further that the belt runs only in one 
direction. 

As an illustration, Fig. 6 shows the arrangement of a belt 
making a quarter turn, the shafts being at right angles. It will 




Fig. 6. 

be noted that the belt in passing from pulley A approaches 
pulley 5 in a direction at right angles to the axis of pulley B, 
and in passing from pulley B it approaches pulley ^ in a direction 
at right angles to the axis of pulley A. It will be noted, too, 
that a plane that is tangent to the rim of one pulley and per- 
pendicular to the axis of the other pulley cuts the other pulley 
in the middle of its rim. 

Non-parallel shafts, whether they intersect or not, may be 
connected by a belt to run in either direction by means of inter- 



TRANSMISSION BY FLAT BELTS AND ROPES 233 

mediate guide pulleys placed on a spindle whose axis is the 
intersection of the middle planes of the principal pulleys. 

13. Rope Gearing. — Ropes running over pulleys having 
F-shaped grooves in their rims are used in preference to belts 
in cases where much power is to be transmitted. The hori- 
zontal distance between the pulley shafts may be as great as 
90 feet, the ropes between the pulleys hanging in catenary 
curves. The materials of which non-metallic ropes are usually 
made are cotton and manila hemp, the former being the better 
on account of its greater flexibility and higher coefficient of 
friction. 

14. Systems of Rope Gearing. — There are two systems of 
rope transmission, known as the multiple and the continuous. 

In the multiple system there is an endless rope for each groove 
of the pulley system, whereas in the continuous system there 
is but one endless rope which, when it leaves the first groove of 
the driving pulley, enters the first groove of the follower pulley. 
Leaving the first groove of the follower, the rope enters the second 
groove of the driver and leaves that groove to enter the second 
groove of the follower, thence to the third groove of the driver, 
and so on until leaving the last groove of the follower, when it 
is directed by a guide pulley to enter the first groove of the 
driver. 

The continuous system is particularly adaptable to cases 
where the distance between the driving and the driven shafts is 
short, and it has the distinct advantage of having but one splice. 
A marked disadvantage lies in the fact that a breakage of the 
rope disables the whole system, which would not be occasioned 
by the breakage of any one of the ropes of a multiple system. 

The ropes most commonly used vary in size from i inch to 
2 inches in diameter, though smaller sizes are used for general 
transmission in manufacturing establishments. 

Each turn in the coil of the rope in the continuous system 



234 THE ELEMENTS OF POWER TRANSMISSION 

has the same tension, so that the driving force is the same as 
that of a multiple system having the same number of separate 
ropes as the continuous system has turns in its coil. 

15. Strength, Weight, and Velocity of Rope Belts. — The 
breaking strength of ropes varies from 7000 to 12,000 pounds 
per square inch of section, but to insure durability it is good 
practice to limit the working stress of a rope to about 150 pounds 
per square inch, indicating a factor of safety of 60. The working 
stress of a rope may be taken as 120^^, in which d is the diameter 
of the rope in inches. 

The weights per linear foot of manila and cotton ropes are 
given very approximately by 0.3^^ and 0.28^^ respectively, 
d being the diameter in inches. 

The velocity of transmission rope varies from 3000 to 6000 
feet per minute, the most efficient speed being 4700 feet. 

16. Frictional Resistance between Rope and Grooved Pulley. 
— The F-shaped grooves in the rims of pulleys used for non- 
metallic rope drives are so dimensioned that the rope presses 
only on the sides of the groove and not on the bottom, thus 
securing a greater resistance to slipping. 

Figure 7 represents a pulley of a rope drive and a cross section 
of the groove and rope. Let a denote the arc of contact of the 
rope and 2 6 the angle of the groove. Let R denote the resultant 
central force due to the normal pressures P and P of the rope 
on the sides of the groove. Consider the small length of arc 
of the rope subtending the angle da, and let Q denote the re- 
sisting pressure at each side of the groove in contact with the 
rope. We have the static equation 

2 Q sin = i? = 2 r sin — = Tda. (See Art. 9, p. 224.) 

The measure of the friction over the small arc considered is 
dT = fiX2Q = fjLX Tda X cosec 6, 



TRANSMISSION BY FLAT BELTS AND ROPES 
dT 



whence 


-— = IX cosec dda. 


Then 


/ ^ = M cosec 6 f 


whence 


T 

log -^ = fia cosec 6, 



235 




Fig. 7. 

in which 6 is half the angle between the sides of the groove. 
As the groove angle is usually about 45 degrees, the value of d 
is 22.5 degrees, and cosec 6 = 2.6. 



230 THE ELEMENTS OF POWER TRANSMISSION 

T 
Then log -7 = 2.6^0;, 

in which 11 is the coefficient of friction and a the arc of contact 
of the smaller pulley. 

Comparing this result with that obtained in Art. 9, p. 224, we 
see that the logarithm of the tension ratio of the grooved pulley 
is 2.6 times as great as that obtained for the flat pulley. 

The coefficient of friction for well lubricated manila ropes lies 
between 0.12 and 0.15, and for cotton ropes between 0.18 and 
0.28. 

T 

The ratio -7 for ropes is much greater than that for flat belts 

owing to the wedging action in the grooves. 

The effect of centrifugal action in ropes is appreciable and 
may be determined in the same manner as for flat belts. 

17. Transmission of Power by Ropes. — As in the case of 
flat belts, the driving force in rope transmission is Ti — T2, and 
the equation for the tension ratio of Art. 9, p. 224, is applicable 
in determining the power transmitted by ropes if we substitute 
IX cosec 6 for /x, d being half the angle of the groove. We shall 
then have 

H.P. per rope = ^-^ ^^j and log ~ = jxa cosec 6 = 2.6 ua, 

550 ^ ^ ^2 

when 2 ^ = 45 degrees, which is commonly the case. 

Example. — How many ropes of 1.125 inches in diameter wiU 
be required to transmit 200 H.P. from a pulley 6 feet in diameter 
making 250 revolutions per minute, the coefficient of friction 
being 0.15 and the arc of contact 150 degrees? 

Solution. — 

Ti = 120^^ = 120 X (1.125)2 = 151.88 pounds. 

Arc of contact = -~r^ = 2.618 radians. 
180 

T 

l0g-;=7 = 2.6 ixa = 2.6 X 0.15 X 2.618 = I.02I02 
T2 



TRANSMISSION BY FLAT BELTS AND ROPES 237 

therefore '^^ = io-5 and T2 = -^ — = 14.46 pounds. 
I2 10.5 

Velocity of ropes = ;: ^ = 78.54 feet per second. 

60 

TT -D (^1 — ^2)^^ (151-88 — 14.46)78.54 . 

H.P. per rope = ^— ^^ -^~ = ^-^ ^ + // — 2^ ^ 10.62. 

550 550 

Number of ropes required = — — = 10.2, say 10. 

19.62 

18. Telodynamic Transmission. — The method of trans- 
mitting power over long distances by means of wire ropes and 
pulleys has been given the name of telodynamic transmission. 
Wire ropes are enormously stronger than ropes made of manila 
and cotton, and when run at high velocities are very efficient 
in the transmission of power. Their excessive wear and great 
cost of replacement have proven to be such serious disadvantages 
as to restrict the employment of a system of transmission other- 
wise admirable. 

The wire ropes commonly used have six strands, each of which 
contains six wires of diameters varying from 0.02 to 0.083 
inch. The strands are wound around a central core of hemp, 
and then the six strands are twisted around the central core of 
the rope, also made of hemp. 

Wire ropes are made of iron and steel, the latter material 
being the better. The working stress of wire rope has been 
fixed at 25,600 pounds per square inch of section, and the weight 
per foot run may be taken as 1.34^^ pounds, in which d is the 
diameter of the rope in inches. 

Owing to the excessive wear which would be occasioned by 
wedging wire ropes in the pulley grooves, as is done with hemp 
and cotton ropes, the grooves are so made as to permit the ropes 
to ride on the bottom. 

The equations found in Arts. 9 and 11 are applicable to wire 
ropes, the coefficient of friction being about 0.24. 



238 THE ELEMENTS OF POWER TRANSMISSION 

PROBLEMS 

1. Two pulleys, 15 inches and 37.5 inches in diameter, are connected by 
a belt. If the 15-inch pulley makes 500 revolutions in 10 minutes, how 
many turns will be made by the other pulley in 25 minutes? Ans. 500. 

2. Sketch an arrangement of four pulleys with belts for driving a fan 
at 1500 revolutions per minute from a shaft making 200 revolutions per 
minute, giving the diameters of the pulleys to be used. 

3. An engine shaft, making n revolutions per minute, carries a 56-inch 
pulley which drives, by means of a belt, a 36-inch pulley on the line shaft. 
The line shaft carries another pulley, 42 inches in diameter, which is belted 
to a 24-inch pulley on a counter shaft. Another pulley on the counter 
shaft is 48 inches in diameter and is belted to a 14-inch pulley on the spindle 
of a dynamo. Find the number of revolutions made in a minute by the 
dynamo spindle. Ans. 9.33^. 

4. The flywheel of an engine is 28 inches in diameter and is belted to 
a pulley 20 inches in diameter on another shaft. A 20-inch pulley on the 
second shaft is belted to a lo-inch pulley on a third shaft, which carries 
an 18-inch pulley, which, in turn, is belted to a 6-inch pulley on the spindle 
of a dynamo. Find the speed of the dynamo when the engine is making 
90 revolutions per minute. If the belt thickness of three-sixteenths inch 
be considered, find the per cent of loss in the speed of the dynamo. 

Ans. 756 revolutions; 3.18 per cent. 

5. The shaft of a high-speed engine, which is making 300 revolutions 
per minute, carries a 14-inch pulley, over which passes a belt to a 20-inch 
pulley on another shaft. A lo-inch pulley on this shaft drives a 20-inch 
pulley on a third shaft carrying a 6-inch pulley which is to be belted to 
the spindle of a machine so that the revolutions of the spindle may be 
45 per minute. Find the diameter of the pulley on the spindle. 

Ans. 14 inches. 

6. A weight of 8000 pounds is suspended from one end of a rope. How 
many turns of the rope must be taken around a circular beam fixed horizon- 
tally in order that a man, who can pull with a force of 250 pounds, may 
keep the rope from slipping, supposing the coefficient of friction to be 0.2? 

Ans. 2.75. 

7. By taking 3 turns of a rope about a post, and holding back with a 
force of 180 pounds, a man just keeps the rope from slipping. The coeffi- 
cient of friction being 0.2, find the weight supported at the other end of the 
rope. Ans. 7810 pounds. 

8. The pulley on an engine shaft is 5 feet in diameter and makes 100 
revolutions per minute. The motion is transmitted from this pulley to 
the main shaft by a belt running on a pulley, the difference in tensions 



TRANSMISSION BY FLAT BELTS AND ROPES 239 

between the tight and slack sides of the belt being 115 pounds. What is 
the work done per minute in overcoming the resistance to motion of the 
main shaft? Ans. 180,714 foot pounds. 

9. A belt having a linear velocity of 350 feet per minute transmits 5 H.P. 
to a pulley. Find the tension in the driving side, supposing it to be double 
that in the slack side. Ans. 943 pounds. 

10. A pulley 3 feet 6 inches in diameter, and making 150 revolutions 
per minute, drives by means of a belt a machine which absorbs 7 H.P. 
What must be the width of the belt so that its greatest tension shall be 
70 pounds per inch of width, it being assumed that the tension in the driving 
side is twice that in the slack side ? Ans. 4 inches. 

11. Find the horse power that may be transmitted by a belt 8 inches 
wide and passing over a 20-inch pulley on the shaft of an engine which 
makes 350 revolutions per minute. The angle at the center subtended by 
the arc of contact of the belt is 160 degrees, and the coefficient of friction is 
0.4. The tension in the driving side may be taken as 80 pounds per inch 
of width, and the stress per square inch of belt section must not exceed 300 
pounds. Find also the thickness of the belt. Ans. 23.94 H.P. ; 0,27 inch. 

12. A belt is to transmit 2 H.P. from a pulley 12 inches in diameter on 
a shaft making 160 revolutions per minute. Find: (i) the tensions in the 
driving and in the slack sides of the belt when the arc of contact is 180 
degrees and the coefiScient of friction is 0.3. (2) The width of the belt 
when the thickness is 0.25 inch, and the safe working stress 320 pounds per 
square inch of belt section. 

Ans. 215.15 pounds; 83.85 pounds; 2.69 inches. 

13. What horse power will a belt 8 inches wide transmit over an 18-inch 
pulley making 300 revolutions per minute, the weight of i square foot 
of the belt being 1.29 pounds, the coefficient of friction 0.3, the arc of con- 
tact of the belt 160 degrees, and the tension per inch of width of the belt" 
not to exceed 80 pounds? Take the weight of a foot of belting i square 
inch in section as 0.43 pound. Ans. 15.8. 

14. A leather belt is required to transmit 2 H.P. from a shaft running 
at 80 revolutions per minute to a shaft running at 160 revolutions per 
minute. Find the stresses in the belt, assuming that the smaller pulley 
is 12 inches in diameter, and that the ratio of the tensions in the tight and 
slack sides of the belt is 2.25 : i. Find also the width of belt, taking the 
working stress at 100 pounds per inch of width. 

Ans. 236.3 pounds; 105 pounds; 2.36 inches. 

15. Find the width of belt necessary to transmit 10 H.P. to a pulley 
12 inches in diameter so that the greatest tension may not exceed 40 pounds 
per inch of width when the pulley makes 1500 revolutions per minute. The 



240 THE ELEMENTS OF POWER TRANSMISSION 

weight of the belt per square foot is 1.5 pounds, the coefficient of friction 
0.25, and the arc of contact 180 degrees. The weight of a cubic inch of 
leather may be taken as 0.036 pound. (The effect of the thickness of the 
belt and of centrifugal action must be taken into consideration.) 

Ans. 8.4 inches. 

16. An 8-inch belt traveling over a 30-inch pulley making 174 revolu- 
tions per minute transmits 18 H.P. The ratio of the tensions in the tight 
and slack sides of the belt is 3.06 : i, the arc of contact 160 degrees, and the 
maximum tension allowed per inch width of belt 80 pounds. Taking the 
velocity of the neutral surface of the belt as the true velocity, it is required 
to find: (i) The coefficient of friction. (2) The thickness of the belt. 

Ans. 0.4; 0.25 inch. 

17. A driving shaft, making 100 revolutions per minute, carries a pulley 
22 inches in diameter, from which a belt communicates motion to a 12-inch 
pulley on a countershaft. On the countershaft is also a cone pulley having 
steps of 8, 6, and 4 inches in diameter, which gives motion to another cone 
pulley, of equal steps, on a lathe spindle. Sketch the arrangement in side 
and end elevations, and find the greatest and least speeds at which the 
lathe spindle can revolve. Ans. 366.66; 91.66. 

18. Determine the horse power that may be transmitted by a belt 
6 inches wide and 0.25 inch thick running at a speed of 60 feet per second. 
The tension in the slack side of the belt is 0.45 of that in the tight side, and 
the maximum allowable stress per square inch of belt section is 280 pounds. 
Taking the weight of a cubic inch of leather as 0.036 pound, to what extent 
does the effect of centrifugal action reduce the power transmitted? 

Ans. 20.87; 17-2 per cent. 

19. It is required to transmit 16 H.P. from a pulley 20 inches in diameter 
by means of a belt which embraces only two-ninths of the circumference 
of the pulley. The thickness of the belt is three-eighths of an inch, the 
•safe working stress is 300 pounds per square inch of belt section, and the 
pulley speed is 120 revolutions per minute. Find the tensions in the two 
parts of the belt, and the width of belt required. 

Ans. 2174.2 pounds; 1333.8 pounds; 19.32 inches. 

20. What horse power will be transmitted from a lo-foot pulley by 12 
ropes of 1.5 inch diameter, the revolutions being 140 per minute, the co- 
efficient of friction 0.14, and the arc of contact 130 degrees? Ans. 306. 

21. A drive of 20 ropes transmits 600 H.P. from a lo-foot pulley mak- 
ing 100 revolutions per minute, the arc of contact of the smaller pulley 
being 160 degrees and the coefficient of friction 0.12. What is the pull in 
each rope? What should be the diameter of the ropes ? 

Ans. 364 pounds; 1.75 inches. 



CHAPTER II 
TRANSMISSION BY TOOTHED WHEELS 

19. Toothed Wheel Gearing. — The usual arrangement of 
toothed wheels in train for the transmission of power is to have 
two wheels of unequal size on each shaft except the first and 
last, making the smaller wheel of a pair on one shaft gear with 
the larger of the pair on the next shaft in the series. The two 
wheels of unequal size on a shaft is the mechanical equivalent 
of a lever with unequal arms, and therefore modifies the power 
that may be transmitted. 

The circles of Fig. 8 represent the pitch circles of two toothed, 
or spur, wheels in gear. The pitch circles of two wheels in gear 
are circles which appear to roll 
upon each other and which pass, 
approximately, through the mid- 
dle of the elevation of the teeth. 
They may be regarded as the 
outlines of two discs which roll 
together by the friction at the 
circumferences. ^^' 

There can be no slipping of one pitch circle over the other, 
owing to the teeth; therefore the same length of circumference 
of each must pass over the point of contact in a given time. 
If D and d represent the diameters of the large and small wheels 
respectively, and N and n the number of their revolutions per 
unit of time, then 

ttDN = Trdn, or DN = dn. 

241 




242 THE ELEMENTS OF POWER TRANSMISSION 

The teeth on the two wheels are the same size, and their 
number will be proportional to the diameters of the wheels, 
and we may write AN = Bn, in which A and B are the numbers 
of teeth of the large and small wheels respectively. 

The pitch of the teeth is the distance measured on the cir- 
cumference of the pitch circle between the centers of two con- 
secutive teeth. The radius of the pitch circle of a spur wheel 
may be found by dividing the product of the pitch and number 
of teeth by 2 TT. 

The formulas for belt pulleys hold for toothed wheels, for the 
belt performs the same ofi&ce as the teeth — it causes the cir- 
cumference of each pulley to move over the same distance in 
the same time. 

With toothed gearing the slipping of belt gearing is avoided 
and an exact velocity ratio may be maintained, provided the 
teeth are carefully constructed on certain geometrical principles. 

20. Arrangement of a Train of Toothed Wheels. — An ar- 
rangement of a train of toothed wheels for the transmission of 
power is shown in Fig. 9, where two wheels of unequal size are 
placed on each axis, except the first and last, and where the 
smaller wheel of any pair gears with the larger wheel of the next 
pair in the train. On the first and last axis there is but one 
wheel. 




Fig. 9. 



In this arrangement for power transmission the driving wheels 
are smaller than the followers, the revolutions of the successive 



TRANSMISSION BY TOOTHED WHEELS 243 

axles decreasing in number. Should a high velocity ratio be 
the object desired, the arrangement would be the reverse of that 
for power transmission, the driving wheels being the larger, 
and, in consequence, the revolutions of the successive axles 
would increase in number. 

Suppose the power to be applied at wheel A. Then wheels 
A, C, and E will be the drivers and wheels B, D, and F the 
followers. Let the letters denoting the wheels denote also the 
numbers of teeth in the wheels respectively. Let iVi, A^2, Nz 
denote the number of revolutions in a period of time of the 
drivers A, C, E respectively; and let fii, n^, fiz be like repre- 
sentations of the followers B, D, F. 

From what has been shown we shall have 

ANi = Bfii, CN2 =Dn2, and ENs = Fuz. 

Multiplying these equations, member by member, we have 

ANi X CN2 X EN^ = Bfii X Dn^ X Fm. 

Since B and C are fixed on the same shaft, we have N2 = Wi; 
and for the same reason A^3 = n2. 

AXCXE fis 



Hence ANiXCxE = BxDxFns, whence 



BXDXF Nv 



That is, the product of the number of teeth in all the drivers, 
divided by the product of the number of teeth in all the fol- 
lowers, is equal to the ratio of the number of revolutions of the 
last wheel to the number of revolutions of the first wheel. This 
ratio is the value of the train, and is denoted by e. 

If, in Fig. 9, A = iS, C = 20, £ = 24, ^ = 36, D = 40, 
F = 72, and Ni = 120, we shall have 

18 X 20 X 24 fis , 

e = — = — ) whence n^ = 10, 

36 X 40 X 72 120 

that is, the last wheel will make 10 revolutions while the first 
wheel is making 1 20 revolutions. 



244 THE ELEMENTS OF POWER TRANSMISSION 

When any number of wheels are in gear, no two of them being 
on the same axis, as in Fig. lo, the combination is the equivalent 
only of a single pair of wheels, viz., the first wheel and the last 
wheel; the intervening wheels simply transfer the motion and 
determine the direction of rotation of the last wheel. If the 
number of idler wheels intervening between the first and last 
wheel be odd, the direction of rotation of the first and last 
wheels will be the same; if even, the rotations will be in opposite 
directions. 




Fig. lo. 

21. Driving End and Load End of Gearing. — Generally 
speaking, the part of a system of gearing to which the motive 
power is applied is called the driving end, and the part at which 
the resistance is overcome, or at which the useful work is done, 
is called the load end. This applies to the operation of all ma- 
chines, and in general terms we have these relations : 

,^ , . . Movement of driving end 

Velocity ratio = —rz -— — — ,° , , 

Movement of load end 

Mechanical advantage = 



Driving force 

^ , , . , ^^ . Useful work performed 

Mechanical efficiency = ^ ^ -, 7-3- • 

Total work expended 

22. Toothed Gearing of Screw-cutting Lathes. — The toothed 
gearing of a screw-cutting lathe is a very important application 
of wheels in train. The driving wheel of the gearing is either 
fast on the lathe spindle or derives motion by means of inter- 



TRANSMISSION BY TOOTHED WHEELS 245 

mediate gearing. In either case the revolutions of the first 
driver are the same as those of the lathe spindle. The inter- 
mediate gearing affords a ready means of throwing the gear 
wheels out of action when the lathe is to be run at high speeds, 
as for polishing; it also enables the direction of rotation of the 
lead screw of the lathe to be changed at will. If, as is usually 
the case, the lead screw is right-handed, the screw to be cut will 
be right-handed or left-handed, according as the direction of 
rotation of the lead screw is the same as, or different from, that 
of the lathe spindle. 

There are two systems of lathe gearing, — simple and compound. 

23. Simple Lathe Gearing. — Fig. 11 is a representation of 
simple lathe gearing, no two wheels being on the same axis. 
Any wheels intervening between the driv- 
ing wheel A and the wheel C on the lead 
screw of the lathe have no influence other 
than to convey the motion, so that the 
only wheels to be considered are the driver 
A and the wheel C on the lead screw. 
If the gearing be such that the lathe spindle 
and the lead screw make the same number 
of turns, the thread cut wil! have the same 
pitch as the thread on the lead screw. 
In all other cases the pitch of the thread 

^ Fig. II. 

to iDe cut will be finer or coarser than 
that on the lead screw in the exact proportion that the revolu- 
tions of the lathe spindle in a unit of time are greater or less 
than those of the lead screw. 

The driving wheel A is fast on the lathe spindle, or so con- 
nected with it as to make the same number of revolutions as 
the spindle. The wheel C is keyed on the lead screw of the 
lathe and the intermediate wheel B is an idler, serving the 
purpose of making the wheels A and C rotate in the same 




246 THE ELEMENTS OF POWER TRANSMISSION 

direction and also of filling in the space between A and C, the 
axis of B being adjustable in a slotted arm. 

The lead screw of the lathe works in a nut on the lathe carriage 
which carries the cutting tool, so that for each revolution of 
the wheel C the cutting tool advances a distance equal to the 
pitch of the lead screw, and it depends entirely upon the ratio 
between the numbers of teeth of wheels C and A as to the number 
of turns the piece of work upon which the thread is to be cut 
will make while the cutting tool is moving through that distance. 
From this we shall have 

Revolutions of lead screw Pitch of screw to be cut 



Revolutions of lathe spindle Pitch of lead screw 

The problem of screw-cutting consists, then, in finding a 
train of wheels in which we shall have 

Pitch of screw to be cut 

c = • 

Pitch of lead screw 

Example. — It is desired to cut a screw of 10 threads to the 
inch with a lathe whose lead screw has a pitch of | inch. Find 
a suitable train of wheels. 

Solution. — 

_ ]^ _ 2 _ 20^ 

" i ~5 "50* 

Giving 20 teeth to the wheel A and 50 teeth to wheel C is one 
of a number of solutions. 

24. Compound Lathe Gearing. — Fig. 1 2 is a representation 
of compound lathe gearing. The two wheels, B and C, of 
different size on the axis intervening between the driver A and 
the wheel D on the lead screw are factors in the value of the 
train. From Art. 20 we have 

A XC 

'~bxd' 



TRANSMISSION BY TOOTHED WHEELS 247 

Example. — It is desired to cut a screw of 16 threads to the 
inch with a lathe whose lead screw has a pitch of | inch. Find 
a suitable train of compound gearing. 




Fig. 12. 
Solution. — 6=-7- = - = -X- = — X — = -=; ;; • 

i ^ 2 ^ 40 g6 B X D 

Hence, referring to Fig. 12, giving 20 teeth to A, and 24, 40, 
and 66 to C, B, and D, respectively, is a probable solution. 

In order to cut a left-hand thread the lathe spindle and the 
lead screw must turn in opposite directions. This may be 
effected by interposing an idle wheel between C and D. 

25. Back Gear Lathe Attachment. — The back gear at- 
tachment to a lathe is a mechanical arrangement to increase 
the power of the lathe at the expense of the speed. The cone 
pulley to which the pinion A is attached, Fig. 13, is loose on 
the lathe spindle. The spur wheel B is keyed to the spindle. 
The spur wheel C and pinion D, carried on the shaft E, form the 
tack gear and can, at will, be thrown into or out of gear with 
A and B. 



248 



THE ELEMENTS OF POWER TRANSMISSION 



The motion of the cone pulley may be conveyed to the lathe 
spindle in two ways: (a) The back gear being disengaged, as 
shown in Fig. 13, the spur wheel B is made to engage with the 
speed cone by means of a bolt, thereby giving to the lathe spindle 
the same revolutions that are made by the cone pulley, (b) 
Disengaging the wheel B from the speed cone and throwing the 
back gear into gear with wheels A and B, the motion of the cone 
pulley is then transmitted to the lathe spindle by means of 
the train ACDB, making the entire system consist of a train 
of which the pulley on the countershaft is the first driver and 
the spur wheel B the last follower. 




Fig. 13. 



26. Transmission of Power by Toothed Wheels. — In the 

different mechanisms employing a train of toothed wheels for 
the transmission of power, either the movement of a small power 
through a comparatively great distance is utiHzed in overcoming 
a much greater resistance through a much smaller distance; or, 
conversely, the movement of a large power through a small dis- 
tance is utihzed in making a smaller resistance move through a 
much greater distance. In the two cases the desired results were 
power and velocity respectively, the underlying mechanical 



TRANSMISSION BY TOOTHED WHEELS 



249 



principle being that, what is gained in power is lost in speed, 
and conversely. 

The power may be applied by hand to the end of a lever, or 
to the crank pin of an engine, the lever or crank to be rigidly 
connected to the first axis of the train. 




Fig. 14. 



Suppose the wheels B and C of Fig. 14 to be fixed on the 
axis c. The driving power P is applied through the wheel A 
tangentially to the pitch surface of the teeth in contact at h, 
the tendency being to turn the wheels B and C in a contra- 
clockwise direction about axis c. The action of P is resisted 
by the reaction R of the load appHed tangentially to the pitch 
surfaces of the teeth of the wheels C and D in contact at d, with 
a tendency to turn the wheds B and C in a clockwise direction 
about the axis c. With these opposite turning tendencies we 
shall have, when motion is about to take place, 

P Xcb = RXcd, 
whence 

P _cd _ Radius of wheel C _ Number of teeth in wheel C 
R ch Radius of wheel B Number of teeth in wheel B 

Example. — The double purchase wheel work of Fig. 15 repre- 
sents a common arrangement appHed to hoisting machinery, 
such as cranes. The numbers attached to the spur wheels and 
pinions indicate the number of teeth they contain. The length 



250 



THE ELEMENTS OF POWER TRANSMISSION 



of the lever handles being i8 inches, the radius of the drum 
lo inches, and the power applied to each of the handles 40 pounds, 
it is required to find: (a) The weight raised at the drum; (b) the 
tangential pressures between the teeth of the wheels; (c) the 
horse power transmitted, supposing the handles to make 24 
turns per minute. 



-'ir^^ — ^P -€ 




1 
^ 



Fig. IS- 

Solution. — Since the radii of wheels are proportional to the 
number of teeth, we may denote by 15 x and 120X the radii of 
the first pinion and the last spur wheel of the train respectively. 

Denoting the length of the power handles and the radius of 
the drum by a and b respectively, we have, by the principle of 
moments, these equations: 

P X a = R1X15X (i) 

Pi X 80 = i?2 X 40 (2) 

P2 X 100 = i?3 X 20 (3) 

PsXi2ox = W Xb (4) 



TRANSMISSION BY TOOTHED WHEELS 251 

Multiplying these equations, member by member, remember- 
ing that Pi = Ri, P2 = R2, and P3 = R3, we have 

Pa X 80 X 100 X 120 = 15 X 40 X 20 X Wb, 

whence 

P _ 15 X 40 X 20 b _ b E.w£P_ _J_ 

W 80 X 100 X 120 a a 80 18 144 

and PF = 11,520 pounds. 

In the above calculations the effect of friction and of the 
diameter of the rope have been neglected. Friction would 
likely reduce the result as much as 30 per cent, and the effective 
drum radius would be the radius of the drum plus the radius 
of the rope. 

The tangential pressures between the surfaces of the teeth 
in contact may be found by means of equations (i), (2), (3), and 
(4) if the pitch of the teeth be known. 

Suppose the pitch of the teeth to be 1.25 inches. Then 

Radius of first pinion = -^-^ ^ = -^ • 

2 TT 8 TT 

From equation (i) we have 

80 X 18 = i^i X ^ ' , whence Pi = 482.5 pounds. 

8 TT 

From equation (2) we have 

482.5 X 80 = P2 X 40, whence P2 = 965 pounds. 
From equation (3) we have 

965 X 100 = P3 X 20, whence P3 = 4825 pounds. 

These results indicate that the teeth of the wheels should be 
made stronger as the drum shaft is approached. This could 
be done by using a coarser pitch for the last pair of wheels in 
gear. 

If Q denotes the tangential pressure at the pitch surface of 



252 THE ELEMENTS OF POWER TRANSMISSION 

any wheel in the train, and V the velocity in feet per minute 
of the wheel at the pitch surface, then 



H.P. transmitted 



QV 



33,000 

and this is constant for any stage of the transmission. 

Thus, if the lever handles make 24 turns per minute we shall 
have at the lever handle axle 

H.P. =-Q^ = 80X2XXI8X.4 ^ „ 8. 

33-000 33^000 X 12 

Since the value of the train is ^V, the last wheel will make 
to = To of a revolution in a minute, and the radius of the last 

wheel is -^-^ = — • 

2 TT TT 

Hence at the drum axle 

H.P. = 4825X2xX7,SX3 ^ ^ g^ 
33,000 X 12 Xtt X 10 

In like manner, the horse power transmitted from axle A to 
axle B may be shown to be the same, thus : 

Axle A makes ^— — - = 4.5 revolutions per minute, and the 
80 

radius of the pinion of 40 teeth is-^^ — = — • Then 

2 TT TT 

H.P. from axle A to axle B = 9^5 X 2. X 25 X 4.5 ^ ^ g. 

33,000 X TT X 12 

In illustration of a combination of belt and toothed wheel 
gears this example is given: 

The motion of an engine shaft is commimicated to a 15-inch 
pulley on a shaft A by means of an open belt passing from a 
36-inch flywheel. The motion of shaft A is transmitted to a 
shaft B by means of spur wheels of J-inch pitch and of 43 and 
24 teeth, an idle wheel intervening. The motion of shaft B is 
commimicated to the spindle of a fan by means of an open 



TRANSMISSION BY TOOTHED WHEELS 253 

belt passing over an 18-inch pulley on shaft ^ to a 12-inch 
pulley on the fan spindle. The engine makes 250 revolutions 
per minute and has a stroke of 12 inches; the mean pressure 
on the crank pin is 90 pounds, and the ratio of the tensions 
in the tight and slack sides of the flywheel belt is 2.25 : i. 
Find: (a) The number of revolutions per minute of the fan; 
(b) the tangential pressure at the point of contact of the teeth 
in gear; (c) the horse power at two or more stages of the trans- 
mission; (d) the difference of the tensions in the tight and slack 
sides of the fan belt. 

Solution. — 

Revolutions of fan per minute = — — — — = 161 2.1;. 

^ 15X24X12 ^ 

Taking moments about the axis of the engine shaft, Fig. 16, 
we have 

18 Ti = 90 X 6 + 18 7*2, whence Ti — T2 = 30. 

Then 2.2$ T2 — T2 = 30, whence T2 = 24, 

and Ti = 30 + 24 = 54 pounds. 

Radius of spur wheel of 43 teeth = ]- — = 5.95, say 6 inches. 

,» 8 X 2 TT 

2 ^o X '^6 
Revolutions of shaft A = — — = 600 per minute. 

15 

To find the tangential pressure, P, at the point of contact, a, 
of the teeth in gear, we take moments about the axis of shaft -4, 
thus, 

7.STi = 6P-h7.sT2, 

whence p^ 7.5^-72) ^7.5X30 ^ 

6 6 ^ "^ ' 

and this pressure is transmitted without change to the point 
of contact, b, the intervening spur wheel having no effect other 



?S4 



THE ELEMENTS OF POWER TRANSMISSION 



than to cause the fan to turn in the same direction as the 
engine. 




H.P. at engine shaft = 



Fig. i6. 

PV _ 90 X 12 TT X 250 
33,000 33^000 X 12 



2.142. 



H.P. at shaft A = ^ — = = 2.142, 

33)Ooo 33)000 X 12 

when considering the 15 -inch pulley on shaft A. 

The horse power being the same at all points in the trans- 
mission, we shall have, when considering the 18-inch pulley on 
shaft B and the tensions in the fan belt, 

H.P. = 2.X42 = ^^^1^^^^20i^^^<fcoX41, 
33,000 X 12 X 24 

whence T/ - T2 = i3-9 pounds. 



TRANSMISSION BY TOOTHED WHEELS 255 

This last result may be obtained by a consideration of the 
equilibrium of the whole system, thus: 

The resistance R between the teeth in contact at a maintains 
with Ti and T2 the equilibrium about the axis of shaft A. The 
contraclockwise moment of P at a is balanced by the clockwise 
moment of R at b, while P slI b maintains with Ti and T2 the 
equihbrium about the axis of shaft B. 

The radius of the wheel of 24 teeth on shaft B being J 

8 X 2 TT 

= 3.34 inches, we shall have, by moments about the axis of shaft By 
9 r/ = 3.34 P + 9 T2\ 

or 9(ri'-r20 =3.34x37.5, 

whence Ti — T2 = i^.g pounds. 

as found above. 

These calculations neglect the losses in the transmission due 

to belt sHpping and friction between the teeth and at the 

bearings, the aggregate of which would probably exceed 30 per 

cent. 

PROBLEMS 

1. A, B, and C are three parallel spindles, A carrying a spur wheel of 
52 teeth which gears with one of 19 teeth on B. On B is another wheel of 
81 teeth gearing with one of 21 -teeth on C. \Vhile A is making 15 turns, 
how many will C make? How many will B make? 

Ans. 158.33; 41.05. 

2. If the change wheels of a lathe have 18, 30, 40, 50, and 88 teeth, show 
an arrangement for cutting a screw of 1 1 threads to the inch, the lead screw 
having a pitch of one-third inch. 

3. With a lead screw of one-half inch pitch, and change wheels of 20, 
24, 30, 40, 55, 60, 80, and 100, show a selection of wheels to cut screws of 
6, II, and 16 threads to the inch. 

4. The back gear of a lathe is in use. Diameter of pulley on counter 
shaft, 4.25 inches; diameter of puUey of lathe, 8.75 inches; pinion on cone 
pulley of lathe, 18 teeth; spur wheel on back shaft, 58 teeth; pinion on back 
shaft, 18 teeth; spur wheel on lathe spindle, 58 teeth. How many revolu- 
tions per minute will the lathe spindle make when the countershaft is mak- 
ing 150 revolutions per minute? Ans. 7.017. 



256 THE ELEMENTS OF POWER TRANSMISSION 

5. Two parallel shafts, whose axes are to be, as nearly as possible, 
30 inches apart, are to be connected by a pair of spur wheels, so that while 
the driver runs at 100 revolutions per minute the follower is required to 
run at only 25 revolutions per minute. Sketch the arrangement, and mark 
on each wheel its pitch diameter and the number of its teeth, the pitch of 
the teeth being 1.25 inches. Determine also the exact distance apart of 
the two shafts. Ans. 29.8295 inches. 

6. Make a sketch of a back gear of a lathe. If the two wheels have 63 
teeth each, and each pinion 25 teeth, find the reduction in the velocity 
ratio of the lathe due to the back gear. Ans. 1 : 6.35. 

7. The double purchase wheelwork of a crane consists of a pinion of 
16 teeth on the handle axle; a wheel and pinion of 64 and 20 teeth respec- 
tively on the first intermediate axle; a wheel and pinion of 80 and 18 teeth 
respectively on the second intermediate axle, and a wheel of 90 teeth on the 
drum axle. The power handles are 20 inches long, the radius of the drum 
II inches, the diameter of the rope 2 inches, and the pitch of the teeth 1.25 
inches. Neglecting friction, it is required to find: (i) The power that must 
be applied at the handles to raise 9600 pounds at the drum; (2) the tan- 
gential pressures between the teeth of the wheels' in gear; (3) the H.P. 
transmitted, supposing the handles to make 20 turns per minute. 

Ans. 72 lbs.; 452.4 lbs.', 1447.7 lbs.; 6434 lbs.; 0.4569 H.P. 

8. The wheelwork of a crane consists of a pinion of 11 teeth gearing 
with a wheel of 92 teeth, and of a pinion of 12 teeth gearing with a wheel 
of 72 teeth on the drum axle. The lever handle being 18 inches long, and 
the radius of the drum 9 inches, it is required to find the ratio of the power 
to the weight raised. Ans. i : 100 nearly. 

9. In the example of a combination of belt and toothed wheel gears 
given at the end of Chapter II, it is required to find the horse power of 
the transmission at shafts A and B by considering the spur wheel of 43 
teeth on shaft A and the one of 24 teeth on shaft B. 



INDEX 



Adhesion of concrete to steel, 204. 
Advantage, mechanical, 244. 
Alloys, 201. 
Angle of shear, 96, 
of twist, 96. 
Applications of center of gravity, 9. 
Area, reduction in, 209. 

Arrangement of a train of toothed wheels, 242. 
Arm of force, 2. 

of couple, 25. 
Assumptions in theory of beams, 67. 
Axis, neutral, 64, 

Back gear lathe attachment, 247. 
Beam, cantilever, 25. 
Beam design, 67. 
Beam, simple, 26. 

Beam resting on three supports, 82. 
uniformly loaded, 34. 

uniformly loaded over part of its length, 45, 48. 
with concentrated load at middle, ss- 
with concentrated moving load, 55. 
with overhanging en^s and uniformly loaded, 42. 
with overhanging ends, uniform and concentrated loads, 51. 
with two equal and symmetrically placed loads, 38. 
Beams, assumptions in theory of, 67. 
continuous, 81. 
deflection of, 73. 

distinction between, and girders, 146. 
examples in deflection of, 75-84. 
examples in resilience of, 109-1 1 1, 
resilience of, 109. 
standard I, 68. 
strength and stiffness of, 85. 
table of relative strength and stiffness of, 87. 
theory of, 62. 
Belts, centrifugal action in, 228. 
flat (see Flat belts). 

257 



258 INDEX 

Belting, leather, coefficient of friction of, 220. 
strength of, 220. 
velocity of, 221. 
Bending moment, 25. 

combined with twisting, 100. 

diagrams, 28. 

first derivative of is the shear, 37. 

influence line of, 58. 
Bending-moment and shear diagrams of cantilevers, 39. 
Bending moments, general case of, 26. 
Bessemer steel, 193. 
Bow's system of lettering, 114. 
Box girder, 68. 

Braced cantilevers, stresses in, 159. 
Brass, 201. 

Brazing, of cast iron, 190. 
Bronzes, 201. 

Cantilever, 25. 

Cantilever beams, bending-moment and shear diagrams of, 39, 133, 134. 

stresses in braced, 159. 
Carbon, graphitic, 185. 

influence of in cast iron, 185. 
Castings, inspection of, 189. 
Cast iron, 184. 

brazing of, 190. 
gray, 185. 

influence of carbon in, 185. 
influence of manganese in, 188. 
influence of phosphorus in, 188. 
influence of silicon in, 187. 
influence of sulphur in, 188. 
malleable, 190. 
mottled, 186. 
shrinkage of, 189. 
strength and hardness of, 189. 
uses in engineering, 189. 
white, 186, 
Center of gravity, 3. 

applications of, 9. 

of arc of circle, 5. 

of a cone, 12. 

of a plane figure, 6. 

of portion of polygon or sector of circle, 10. 

of semicircular arc, 13. 

of a system, 4. 



INDEX 259 



Center of gravity, of a triangle, 9. 
Center of gyration (see Radius of gyration). 
Center of moments, i. 
Centrifugal action in belts, 228. 
CoefiScient of elasticity, 62. 

friction in leather belting, 220. 

rigidity, 63. 
Columns, 89. 

design of, 91. 
Combined torsion and bending, 100. 
Compositions, 201. 
Compression tests, 217. 
Concrete, 202. 

reinforced, 202. 
Cone, center of gravity of, 12. 
Continuous beams, 81. 
Contrary flexure, points of, 74. 
Copper, 199. 
Couple, 25. 

arm of, 25. 
Crane, framed, 177. 
Crucible steel, 193. 

Dangerous section, 36. 
Deficient frames, 179. 
Deflection of beams, 73, 

examples in, 75-84. 
Deformation, internal work due to, 105. 
Design of columns, 91. 
Diagram, force, 116. •» 

frame, 148. 

of bending moments, 28. 

reciprocal, 148. 

shear, 29. 

stress, 148. 

stress-strain, 212. 
Driving end of gearing, 244. 
Ductility, 212. 

Efficiency, mechanical, 244. 
Elastic limit, 208, 213. 
Elasticity, coefiicient of, 62. 

modulus of, 62. 

transverse, 63. 
Elongation, 208. 
Engineering materials, 184. 



26o INDEX 

Equilibrium, 2. 

Equivalent twisting moment, 100. 

Extensometer, 214. 

Factor of safety, 208. 
Ferromanganese, 188. 
Fink truss, 165. 
Flanges of I beams, 68. 
Flat belt gearing, 219. 

and pulley, frictional resistance between, 224. 
connections between non-parallel shafts, 231. 
transmission of power by, 227. 
transmission, velocity ratio in, 221. 
Flat belts, materials for, 219. 
slip of, 222. 
tensions in, 223. 

transmission of power by, 219, 227. 
Force action at a joint, 147. 
Force diagram, 116. 

pole of, 119. 
vectors of, 119. 
Forces, supporting, 146, 148. 
Forms of test specimens, 210. 
Frame, 145. 

diagram, 148. 

force action at joint of, 147. 
Framed crane, 177. 

structures, 145. 
Frames, deficient, 179. 

redundant, 179. 
Frictional resistance between flat belt and pulley, 2 24. 

rope and grooved pulley, 234. 
Funicular polygon, 117. 

a bending-moment diagram, 125. 

examples in its application to beams, 126-143. 

illustrations of, 120. 

Galvanizing, 201. 

Gearing, compound lathe, 246. 

driving end of, 244. 

flat belt, 219. 

load end of, 244. 

rope, 233. 

simple lathe, 245. 

toothed wheel, 241. 
Girder, box, 68. 



INDEX 261 

Girder, Linville or N, 164. 

Warren, 162. 
Girders, distinction between beams and, 146. 
Graphic statics, 114. 

Gravity, center of (see Center of gravity). 
Gun metals, 201. 
Gyration, center of (see Radius of gyration). 

Hooke's law, 63. 

I beams, standard, 68. 

flanges and web of, 68. 
Illustrations of moments of inertia and radii of gyration, 19-22. 
Inertia, moments of about parallel axes, relation between, 18. 
Inertia, plane moment of, 17. 

circular surface, 21. 

parallelogram, 20. 
Inertia, polar moment of, 17. 

right circular cone, 22. 
Inflection, points of, 74. 
Influence line of bending moment, 58. 

shear, 60. 
Internal work due to deformation, 105. 
Iron, cast (see Cast iron). 
Iron, wrought (see Wrought iron). 

Lathe, back gear attachment of, 247. 
compound gearing of, 246. 
simple gearing of, 245. 
Lathes, screw cutting, toothed gearing of, 244. 
Lead, 200. 

Leather belting, coefficient of friction of, 220. 
strength of, 220. 
velocity of, 221. 
weight of, 220. 
Lettering, Bow's system of, 114. 
Limit, elastic, 208, 213. 
Linville or N girder, 164. 
Load end of gearing, 244. 
Loads, dead, 145. 
live, 145. 
moving, 55. 
on structures, 145. 
suddenly applied, effect of, 105. 

Machines, testing, 209. 
Manganese, influence in cast iron, 188. 
steel, 196. 



262 INDEX 

Materials, engineering, 184. 

average physical properties of, 217. 
diflferent kinds of tests of, 207. 
Materials for flat belts, 219. 
Mechanical advantage, 244. 
Mechanical efficiency, 244. 
Modulus of elasticity, 62. 
resilience, 107. 
rigidity, 63. 
section, 66. 

polar, 98. 
Moment, i. 

equivalent twisting, 100. 
resisting, 65, 97. 
twisting, 95. 
Moment of inertia or second moment (see Inertia). 
Moments, bending (see Bending moments), 
center of, i. 
clockwise, 2. 
contraclockwise, 2. 
Moving loads, 5$. 

N or Linville girder, 164. 
Neutral axis, 64. 

surface, 64. 
Nickel steel, 199. 

Open-hearth steel, 194. 

Permanent set, 213. 

Phosphorus, influence of in cast iron, 188. 

steel, 196. 
Plasticity, 212. 
Point, yield, 213. 

Points of inflection or of contrary flexure, 74. 
Polar moment of inertia, 17. 
Pole of force diagram, 119. 
Polygon, funicular, 117. 

examples in its application, 120. 
Power, transmission of by flat belts, 227. 
ropes, 236. 
shafts, 100. 
toothed wheels, 248. 

Radius of gyration, plane, 19. 

of circular surface, 21. 
of parallelogram, 20. 



INDEX 263 

Radius of gyration, polar, of right circular cone, 22. 

Rankine's formula for columns, 89. 

Ratio, velocity, 244. 

Reactions, support, 26, 148, 149. 

Reciprocal diagram, 148. 

drawing of, 151. 

examples in drawing of, 152-156. 
Reduction in area, 209. 
Redundant frames, 179. 
Reinforced concrete, 202, 

proportion of reinforcement in, 205. 
Resilience, 105. 

modulus of, 107. 
of beams, 109. 
of torsion, 112. 
Resisting moment for bending, 65. 

torsion, 97. 
Riehle testing machine, 209. 
Riehle-Yale extensometer, 214. 
Rigidity, coefficient of, 63. 
Rolling loads, 55. 
Roof truss fixed at the ends and with wind pressure, 167. 

one end, free at the other, wind and dead loads, 173. 
Rope and grooved pulley, frictional resistance between, 234. 
Rope belts, strength, weight and velocity of, 234. 
Rope gearing, 233. 

systems of, 233. 
Ropes, transmission of power by, 236. 

Safety, factor of, 208. 

Screw cutting lathes, compoimd gearing of, 246. 
simple gearing of, 245. 
toothed gearing of, 244. 
Second moment, 18. 
Section, dangerous, 36. 

method of determining stresses, 157. 
plane modulus of, 66. 
polar modulus of, 97. 
Semicircular arc, center of gravity of, 13. 
Set, permanent, 213. 
Shafts, 95. 

transmission of power by, 100. 
Shear, 29, 

angle of, 96. 
diagrams, 29. 
influence line of, 60. 



264 INDEX 

Silicon, influence in cast iron, 187. 

steel, 196. 
Simple lathe gearing, 245. 
Slip in flat belts, 222. 
Specimens, forms of test, 210. 
Spelter, 201. 
Spiegeleisen, 188. 
Statics (see Graphic statics). 
Steel, 192. 

adhesion of concrete to, 204. 

Bessemer process of making, 193. 

Bessemer and open-hearth compared, 195. 

cement or blister, 193. 

crucible, 193. 

influence of manganese in, 196. 

influence of phosphorus in, 196. 

influence of silicon in, 196. 

influence of sulphur in, 195, 

nickel, 199. 

semi, 196. 

Siemens-Martin or open-hearth process, 194. 

structural, 193. 

tempering of, 198. 

tension test of, 216. 

tungsten, 199. 

uses in engineering, 199. 
Steels, special, 198. 
Strain, 62, 207. 

Strength and stiffness of beams, 85. 
of leather belting, 220. 
of rope belts, 234. 
ultimate, 208. 
Stress, 62, 207. 

kinds of, 207. 
or reciprocal diagram, 148. 
rule for determining kind of, 156. 
section method of determining, 157. 
Stress-strain diagram, 212. 
Stresses in braced cantilevers, 159. 
Structural steel, 193. 
Structures, framed, 145. 
Strut, 145. 

Suddenly applied loads, effect of, 105. 
Sulphur, influence of in cast iron, 188. 

steel, 195. 
Support reactions, 26, 148, 149. 



INDEX 265 

Supporting forces, 146, 148. 
Surface, neutral, 64. , 
System of lettering, Bow's, 114. 
Systems of rope gearing, 233. 

Telodynamic transmission, 237. 
Tension test of steel, 216, 
Tensions in flat belts, 223. 
Test specimens, forms of, 210. 

compression, 217. 
Tests, different kinds of, 207. 
Testing machines, 209. 
Tie, 145. 
Timber, 202. 
Tin, 201. 
Tin plate, 201. 

Toothed gearing of screw-cutting lathes, 244. 
Toothed wheel gearing, 241. 
Toothed wheels, arrangement of train of, 242. 
transmission of power by, 248. 
value of train of, 243. 
Torque, 95. 
Torsion, 95. 

combined with bending, 100. 
resilience of, 112. 
Transmission of power by flat belts, 227. 
ropes, 236. 
shafts, 100. 
toothed wheels, 248. 
Transmission, telodynamic, 237. 
Transverse elasticity, 63. 
Truss, Fink, 165. 

principal rafter of, 146. 

roof, fixed at one end, free at the other, wind and dead loads, 173. 
roof, fixed at the ends, with wind pressure, 167. 
Trusses, 146. 

bays or panels of, 146. 
braces of, 146. 
chord members of, 146. 
web members of, 146. 
Tungsten steel, 199. 
Twist, angle of, 96. 
Twisting moment, 95. 

equivalent, 100. 

Ultimate strength, 208. 



266 INDEX 

Vectors of force diagram, 119. 
Velocity of flat belts, 221. 
rope belts, 234. 
Velocity ratio, 244. 

in flat belt transmission, 221. 

Warren girder, 162, 
Web of I beam, 68. 
Weight of cotton belting, 220. 

leather belting, 220. 

rope belting, 234. 

rubber belting, 220. 
Wind pressure, 150. 

table of, 150. 

Yield point, 213. 

Zinc, 201. 










■ ■ ■::■ l~ ■ .■.-Cfi,:-.'. 






^il 














